Find the partial derivative of $f$ in terms of $g$ and $h$ of $f(x,y)=g(x)h(y)$
My attempt:
By the chain rule $\partial_1f(x,y)= \partial_1f(g,h)\partial_1g(x)+\partial_2f(g,h)\partial_1h(y)$. Now $h(y)$ is a constant in respect to $x$, so $\partial_1f(x,y)=\partial_1f(g,h)\partial_1g(x)$
$$\partial_1f(g,h)=\lim\limits_{k \to 0}\frac{f(g+k,h)-f(g,h)}{k}$$ $$=\lim\limits_{k \to 0}\frac{g(x+k)h(y)-g(x)h(y)}{k}$$ $$=h(y)\lim\limits_{k \to 0}\frac{g(x+k)-g(x)}{k}$$ $$=h(y)\partial_1g(x)$$
Your final answer is correct, but your notation is not very precise. For instance, you write $f(g+k,h)$ at some point. But $k\in\mathbb{R}$ while $g$ is a function, so it doesn't make sense to add them. Apart from the equality $$ \partial_1f(g,h)=\lim\limits_{k \to 0}\frac{f(g+k,h)-f(g,h)}{k}, $$ which does not make much sense for the reason mentioned above, everything else is right.
I would have simply shown it as follows:
$$\frac{\partial f}{\partial x}=\frac{\partial g}{\partial x}h+\frac{\partial h}{\partial x}g=\frac{\partial g}{\partial x}h,$$ where I used that fact that $\frac{\partial h}{\partial x}=0$ since $h$ does not depends on $x$.
Similarly you have
$$\frac{\partial f}{\partial y}=\frac{\partial g}{\partial y}h+\frac{\partial h}{\partial y}g=\frac{\partial h}{\partial y}g,$$