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No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a,t\in\mathbb R$, and we also set $g(x,0)=0$.

The function is still undefined on the $y$ axis. Note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$). Hence setting $g(0,b)=0$ will keep $g$ continuous. Moreover, all directional derivatives exist since $g$ is linear on lines through the origin. Finally, the same procedure works for $h$.

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a\in\mathbb R$, $-1\le t\le 1$, and we also set $g(x,0)=g(0,y)=0$.

So far, $g$ has been defined on the closed set $\{ |y|\ge x^2\}\cup \{ (x,0)\}$, and it is continuous there. To verify continuity on the $y$ axis, note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$).

The existence of all directional derivatives at the origin is already guaranteed because the current domain contains a line segment in every direction, and $g$ is linear there. Finally, we extend $g$ to $\{ 0<|y|<x^2\}$. We can make $g$ smooth away from the origin.

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a,t\in\mathbb R$, and we also set $g(x,0)=0$.

The function is still undefined on the $y$ axis. Note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$). Hence setting $g(0,b)=0$ will keep $g$ continuous. Moreover, all directional derivatives exist since $g$ is linear on lines through the origin. Finally, the same procedure works for $h$.

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a,t\in\mathbb R$, and we also set $g(x,0)=0$.

The function is still undefined on the $y$ axis. Note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$). Hence setting $g(0,b)=0$ will keep $g$ continuous. Moreover, all directional derivatives exist since $g$ is linear on lines through the origin. Finally, the same procedure works for $h$.

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a,t\in\mathbb R$, and we also set $g(x,0)=0$. This function is continuous and all directional derivatives exist since $g$ is linear on lines through the origin. Of course, the same procedure works for $h$.

No. Consider $x'=g(x,y)$, $y'=h(x,y)$. If we take take $g(x,y)=2|x|^{1/2}$ for $y=x^2$ similarly $h=4|x|^{3/2}$ on $y=x^2$, then we can check directly that $x=t^2$, $y=t^4$, $t\ge 0$, and $x=y=0$ are solutions with initial value $(0,0)$, independently of how we define $g,h$ off the parabola.

To define $g,h$ everywhere, we interpolate linearly, starting from the value on the parabola, by letting $g(ta,ta^2)=tg(a,a^2)$, $a,t\in\mathbb R$, and we also set $g(x,0)=0$.

The function is still undefined on the $y$ axis. Note that if $(ta,ta^2)\to (0,b)$ with $b\not= 0$, then $|a|\to\infty$, so $tg(a,a^2)=2t|a|^{1/2}\to 0$ (and also $t|a|^{3/2}\to 0$, which we need in the corresponding step for $h$). Hence setting $g(0,b)=0$ will keep $g$ continuous. Moreover, all directional derivatives exist since $g$ is linear on lines through the origin. Finally, the same procedure works for $h$.