Can I compute $\frac{\partial g}{\partial x}$ to be $\frac{df}{dx}$? The reason is I think $\mathit{f}$ is a one variable function. So $$\frac{\partial g}{\partial x}=\frac{df}{dx}=\frac{1}{y}f'$$ and $$\frac{\partial g}{\partial x}=\frac{df}{dy} = \frac{-x}{y^2}f'$$
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What does $g\binom{x}{y}$ means? I think that is $g(x,y)$. Then use Euler's theorem of homogeneous function differentiation. See, $g(tx, ty)=t^{0}g(x,y)$. – Alapan Das May 18 '20 at 05:18
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@Alapan Das Thanks, $$g\binom{x}{y}$$ means g(x, y). I will search about this theorem. – Steven Lu May 18 '20 at 05:31
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@Alapan Das I have a question. In this case, if $f: \mathbb{R} \to \mathbb{R}$ is differentiable , can I say $g: \mathbb{R}^2 \to \mathbb{R}$ is also differentiable? – Steven Lu May 18 '20 at 05:54
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Your solution is alright, you can interchange the two, as when you take partial derivatives, you are assuming the other variable is constant – Dhanvi Sreenivasan May 18 '20 at 06:03