Let $A \subseteq \mathbb R^2$ be open and $f : A \to \mathbb R$ be a function whose all partial derivatives exist in A, and let $g = (g_1,g_2) : \mathbb R \to A$ be a differentiable function. Does the function $h : \mathbb R \to \mathbb R, h := f\circ g$ have a derivative, and if it does what is it.
Attempt: So if we assume $f$ is also differentiable, then we get (from chain rule) that $$h'(t)=\frac{\partial f}{\partial x}(g(t))g_1'(t)+\frac{\partial f}{\partial y}(g(t))g_2'(t),\quad t\in \mathbb R$$ So i assume that if there exists a derivative of $h$ in $t\in \mathbb R$ it is equal to that. Problem is, i can't prove that it exists since $f$ doesn't have to be differentiable.
I claim that $h$ is not necessarily differentiable. To see this, consider the functions $$f\colon\mathbb{R}^{2}\to\mathbb{R},\qquad f(x,y):=\begin{cases}0&(x,y)=(0,0)\\2xy/(x^{2}+y^{2})&(x,y)\neq(0,0)\end{cases}$$ and $$g\colon\mathbb{R}\to\mathbb{R}^{2},\qquad g(t):=(t,t).$$ Then $f$ is partially differentiable at $(0,0)$. Namely, if $e_{1}=(1,0)$ and $e_{2}:=(0,1)$, then $$\frac{f(\xi e_{i})-f(0,0)}{\xi}=0\to0\qquad\text{as}\qquad\xi\to0\qquad\text{for}\qquad i\in\{1,2\}.$$ However, $f$ is not differentiable at $0$ (in fact not even continuous, check this!). Clearly $g$ is differentiable. Finally, $h:=f\circ g$ is not differentiable in $0$. Indeed, for $t\in\mathbb{R}$ we have $$h(t)=\begin{cases}0&t=0\\1&t\neq0\end{cases}.$$