Partial of $f(x,y)$ wrt $y = 0$ implies $f(x,y) = h(x)$

Question

Prove if $dyf = 0$ for all points in $R2$ then $f(x, y) = h(x)$

I am struggling to articulate this formally though it seems almost obvious. If there were a $y$ term in $f$ then $dyf$ clearly could not be $0$.

I tried defining for a given $x$, $gx(y) = f(x, y)$ and since $gx'(y) = dyf = 0$ then $gx(y)$ must be constant regardless of $x$ but I am not certain how to proceed

Thanks!

Answer 1

Use the fundamental theorem of calculus to integrate the derivative of $f$ with respect to $y$.

Spoiler:

$$\frac{\partial f}{\partial y}(x,y)=0\implies \int \frac{\partial f}{\partial y}(x,y)~dy=\int 0~dy\implies f(x,y)=h(x)$$ for some function $h(x)$ which does not depend on $y$.