Suppose I have a function $g(y,x)$ which is differentiable with respect to both arguments. I know that $y=f(x)$ is bijective, and thus the inverse function exists $x = f^{-1}(y)$. My question is when I calculate $$\frac{d g(y,x)}{d y}$$ should I care about $x$ as well because $x$ implicitly changes when $y$ changes?
I know that when I calculate $$\frac{d g(y,x)}{d x} $$ I have to take into account the fact that $y=f(x)$ also changes in $x$
Yes. $$\frac{\frac{\text dg}{\text dx}}{\frac{\text dg}{\text dy}} = \frac{\text dy}{\text dx} = f'(x) \Rightarrow \frac{\text dg}{\text dy} = \frac{1}{f'(x)} \frac{\text dg}{\text dx}$$
So, if you calculate $\frac{\text dg}{\text dx}$ by taking $f(x)$ into account, you can easily calculate $\frac{\text dg}{\text dy}$ with this relationship.