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The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial x^\ell}\left(\mathrm{d} x^\ell - z\,\mathrm{d}y^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-y^\ell\frac{\partial z}{\partial x^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial x^\ell}\,\mathrm{d}\left(x^\ell - z\,y^\ell\right). \tag3 $$ Now, defining functions $w^\ell = x^\ell - z\,y^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z = \bar z$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now $\Phi$ maps the hyperplane $z=\bar z$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs, one the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + y\cdot y}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - (x\cdot y)^2\,}}{1 + y\cdot y}, $$ and the other the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering a more generic hyperquadric with no real points, though this is far from exhausting all of the global solutions.)

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial y^\ell}\left(\mathrm{d} y^\ell - z\,\mathrm{d}x^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-x^\ell\frac{\partial z}{\partial y^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial y^\ell}\,\mathrm{d}\left(y^\ell - z\,x^\ell\right). \tag3 $$ Now, defining functions $w^\ell = y^\ell - z\,x^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z - \bar z=0$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now, $\Phi$ maps the hyperplane $z-\bar z=0$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs: One is the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + x\cdot x}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - (x\cdot y)^2\,}}{1 + x\cdot x}, $$ and the other is the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering more generic hyperquadrics with no real points, though this is far from exhausting all of the global solutions.)

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial x^\ell}\left(\mathrm{d} x^\ell - z\,\mathrm{d}y^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-y^\ell\frac{\partial z}{\partial x^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial x^\ell}\,\mathrm{d}\left(x^\ell - z\,y^\ell\right). \tag3 $$ Now, defining functions $w^\ell = x^\ell - z\,y^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z = \bar z$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now $\Phi$ maps the hyperplane $z=\bar z$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs, one the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + y\cdot y}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - x\cdot y\,}}{1 + y\cdot y}, $$ and the other the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering a more generic hyperquadric with no real points, though this is far from exhausting all of the global solutions.)

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial x^\ell}\left(\mathrm{d} x^\ell - z\,\mathrm{d}y^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-y^\ell\frac{\partial z}{\partial x^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial x^\ell}\,\mathrm{d}\left(x^\ell - z\,y^\ell\right). \tag3 $$ Now, defining functions $w^\ell = x^\ell - z\,y^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z = \bar z$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now $\Phi$ maps the hyperplane $z=\bar z$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs, one the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + y\cdot y}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - (x\cdot y)^2\,}}{1 + y\cdot y}, $$ and the other the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering a more generic hyperquadric with no real points, though this is far from exhausting all of the global solutions.)

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial x^\ell}\left(\mathrm{d} x^\ell - z\,\mathrm{d}y^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-y^\ell\frac{\partial z}{\partial x^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial x^\ell}\,\mathrm{d}\left(x^\ell - z\,y^\ell\right). \tag3 $$ Now, defining functions $w^\ell = x^\ell - z\,y^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z = \bar z$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now $\Phi$ maps the hyperplane $z=\bar z$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric in $\mathbb{C}^{n+1}$ $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image under $\Phi$ is the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + y\cdot y}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - x\cdot y\,}}{1 + y\cdot y}. $$ Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering a more generic hyperquadric with no real points, though this is far from exhausting all of the global solutions.)

The answer is "Yes, there are many solutions with $u$ and $v$ non-constant."

Here is one way to understand this problem: First, set $z = u+iv$ and note that the above equations become the complex system $$ \frac{\partial z}{\partial x^\ell} + z\,\frac{\partial z}{\partial y^\ell} = 0, \qquad \ell = 1,\ldots,n. \tag1 $$ Written in differential forms, this becomes (summation on $\ell$ implied) $$ 0 = \mathrm{d}z - \frac{\partial z}{\partial x^\ell}\left(\mathrm{d} x^\ell - z\,\mathrm{d}y^\ell\right),\tag2 $$ which can be written in the form $$ 0 = \left(1-y^\ell\frac{\partial z}{\partial x^\ell}\right)\,\mathrm{d}z - \frac{\partial z}{\partial x^\ell}\,\mathrm{d}\left(x^\ell - z\,y^\ell\right). \tag3 $$ Now, defining functions $w^\ell = x^\ell - z\,y^\ell$ on $\mathbb{R}^{2n}\times \mathbb{C}$ (where $z$ is regarded as the coordinate on the $\mathbb{C}$-factor), one has that the mapping $\Phi = (w^1,\ldots,w^n,z):\mathbb{R}^{2n}\times \mathbb{C}\to\mathbb{C}^{n+1}$ is a complex coordinate chart away from the (real) hyperplane $z = \bar z$ in $\mathbb{R}^{2n}\times \mathbb{C}$. Moreover, the above interpretation (3) shows that, if $z(x,y) = u(x,y)+i\,v(x,y)$ where $v(x,y)$ is nowhere vanishing, then $\Phi$ maps the graph of $z$ in $\mathbb{R}^{2n}\times \mathbb{C}$ to a holomorphic hypersurface in $\mathbb{C}^{n+1}$ if and only if $z(x,y)$ satisfies (1).

Now $\Phi$ maps the hyperplane $z=\bar z$ into $\mathbb{R}^{n+1}\subset\mathbb{C}^{n+1}$, so if one starts with a holomorphic hypersurface $Q\subset\mathbb{C}^{n+1}$ that is disjoint from $\mathbb{R}^{n+1}$ and can show that its pre-image under $\Phi$ is the graph of a function $z:\mathbb{R}^{2n}\to\mathbb{C}$, then that function $z= u + i\,v$ will be a solution to the equations (1).

As an example, consider the (holomorphic) hyperquadric $Q\subset\mathbb{C}^{n+1}$ defined by $$ (w^1)^2 + \cdots + (w^n)^2 + z^2 + 1 = 0, $$ which does not meet $\mathbb{R}^{n+1}$. It is easy now to check that the pre-image $\Phi^{-1}(Q)\subset \mathbb{R}^{2n}\times\mathbb{C}$ is the union of two graphs, one the graph of the function $z(x,y) = u(x,y) + i\,v(x,y)$ where $$ u(x,y) = \frac{x\cdot y}{1 + y\cdot y}\quad\text{and}\quad v(x,y) = \frac{\sqrt{(1 + x\cdot x)(1 + y\cdot y) - x\cdot y\,}}{1 + y\cdot y}, $$ and the other the graph of the function ${\bar z}(x,y) = u(x,y) - i\,v(x,y)$. Thus, this is an example of a non-constant solution to the original equations. (Examples with less symmetry can be constructed by considering a more generic hyperquadric with no real points, though this is far from exhausting all of the global solutions.)