Putnam 2009 A6

Let $f:[0,1]^2 \to \mathbb{R}$ be a continuous function on the closed unit square such that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ exist and are continuous on the interior $(0,1)^2$. Let $a = \int_0^1 f(0,y) dy$, $b = \int_0^1 f(1,y) dy$, $c = \int_0^1 f(x,0) dx$, $d = \int_0^1 f(x,1) dx$. Prove or disprove: There must be a point $(x_0,y_0)$ in $(0,1)^2$ such that $$\frac{\partial f}{\partial x} (x_0,y_0) = b - a \text{and} \frac{\partial f}{\partial y} (x_0,y_0) = d - c.$$