Let $f:\mathbb R^2\rightarrow \mathbb R$ has first partial derivatives and $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}=0, \text{for all } (x,y)\in\mathbb R^2$. Show that $f$ is constant. (Hint: Show that the restriction of $f$ to a line parallel to one of the coordinate axes is constant).
My attempt: Following the hint, consider the restriction of $f$ to a line $y=c$, then $\frac{\partial f}{\partial x} (x,c) = f_x(x,c)=0$. Pick 2 points a and b, then by mean value theorem there is $x_0$, such that $f_x(x,c) = \frac{f(b,c)-f(a,c)}{b-a}=0 \implies f(b,c)-f(a,c)=0 \implies f(b,c)=f(a,c)\implies f(x,c)=const.$
We can show the same way that $f(c,y)=const$.
Am I thinking in the right direction? If so, can I just combine $f(x,c)=const$ and $f(c,y)=const$ to get $f(x,y)=const$? Did I miss something?
Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that $$f(x_1,y_1) = f(x_2,y_1) \quad \text{and} \quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.