What if the intermediate point is $c=(0,0)$ in the MVT? Let $f: \Bbb R ^2 \to \Bbb R$ be differentiable everywhere. Assume $f(-\sqrt 2, -\sqrt 2) = 0$ and also that
  
  $$ \left| \frac {\partial f} {\partial x} (x,y) \right| \le | \sin (x^2 + y^2) | \\
\left| \frac {\partial f} {\partial y} (x,y) \right| \le | \cos (x^2 + y^2) |$$
  
  for each $(x,y) \in \Bbb R ^2 \setminus \{ (0,0) \}$. Prove that $|f(\sqrt 2, \sqrt 2)| \le 4$.


So by Mean Value Theorem there is a point $c$ in the straight line segment that joins $(\sqrt{2},\sqrt{2})$ and $(-\sqrt{2},-\sqrt{2})$ such that,

$\|f(\sqrt{2},\sqrt{2})-f(-\sqrt{2},-\sqrt{2})\|\leq |\nabla f(c)| \;\|(\sqrt{2},\sqrt{2})-(-\sqrt{2},-\sqrt{2})\|$.

This implies $\|f(\sqrt{2},\sqrt{2})\|\leq |\nabla f(c)| 4$. Now if $c\neq (0,0)$ then by the given inequalities we can show the result. But, what if $c=(0,0)$? Because $(0,0)$ is a point on the line segment.
Write

$$f(\sqrt 2,\sqrt 2) = f(\sqrt 2,\sqrt 2) - f(0,0) + f(0,0) - f(-\sqrt 2,-\sqrt 2).$$

Use the mean value theorem on each segment; recall the MVT doesn't require differentiability at the end points.
The problem is $f$ may be discontinuous at $0$.

However we can evade $(0,0)$ and follow two line segments.

Call A = $(\sqrt{2},\sqrt{2})$.

Fix $\epsilon&gt;0$ and call E = $(0, \epsilon)$.

Take the path -A to E to +A. Then by making $\epsilon$ arbitrarily small we can approach 4 as closely as needed. Hence $f(A)$ must be $\le 4$.