Consider a continuously differentiable function $f : \mathbb{R}^{n} \to \mathbb{R}^{n}$ such that $f(0) = 0$ and $\langle f(x), x \rangle \ge 0$ for all $x \in \mathbb{R}^{n}$. Does there exist a continuous function $A : \mathbb{R}^{n} \to \mathbb{R}^{n \times n}$ such that $A(x)x = f(x)$ and $\langle A(x) y, y \rangle \ge 0$ for all $x,y \in \mathbb{R}^{n}$?

It is always possible to write such a function $f$ in terms of a continuous matrix function $A$. In particular,
\begin{equation}
A(x) := \int_{0}^{1} \nabla f(sx) \, ds.
\end{equation}
But this matrix is not necessarily positive semidefinite. $A$ is unique up to an additive continuous matrix function $L$ satisfying $L(x)x = 0$ for all $x \in \mathbb{R}^{n}$ since $\big( A(x) + L(x) \big)x = A(x)x = f(x) $.

Edit: Let me elaborate further and also address diracdeltafunk's
comment. On $\mathbb{R}$ the problem at hand is straightforward. The desired function $a$ exists if and only if $f(x)/x$ is continuous at $x = 0$ (which it is if $f$ is $C^{1}$, and the above integral formula will then give the unique answer). 

In $\mathbb{R}^{2}$, take for instance $f(x) = (x_{2}^{2}, -x_{1}x_{2} )$. Then $\langle f(x) , x \rangle = 0 $ for all $x \in \mathbb{R}^{2}$. However, using the above integral formula,
\begin{equation}
A(x) = \begin{pmatrix}
0 & x_{2} \\ -\tfrac{1}{2} x_{2} & -\tfrac{1}{2} x_{1}
\end{pmatrix}
\end{equation}
which is not P.S.D. But we can choose
\begin{equation}
A(x) = \begin{pmatrix}
0 & x_{2} \\ -  x_{2} & 0
\end{pmatrix}
\end{equation}
which is P.S.D.