Recall the definition of the partial derivatives:
$\partial_x u(x_0,y_0)=\lim_{h\to 0}\frac{u(x_0+h,y_0)-u(x_0,y_0)}{h},$ and similarly for $\partial_y u$. If you restrict $u$ exclusively to the line $y=x$, then this difference quotient cannot be found. Even if $x_0=y_0$, notice that for all $h\neq 0$, $x_0+h\neq x_0$. Therefore the correct computation of the partial derivative $\partial_x u(x_0,x_0)$ is
$ \begin{align*} \partial_x u(x_0,x_0)&=\lim_{h\to 0}\frac{u(x_0+h,x_0)-u(x_0,x_0)}{h}\\ &=\lim_{h\to 0}\frac{(x_0+h)^2+x_0^2-2x_0^2}{h}\\ &=\lim_{h\to 0}\frac{2x_0h+h^2}{h}\\ &=2x_0, \end{align*}$
and similarly for $\partial_yu(x_0,x_0)$.
You could consider the single variable function $f(x):=u(x,x)$, but that is another story. Its derivative tells you how $u$ changes along the line $y=x$. In fact, $f'(x)$ is $\sqrt 2$ times the directional derivative of $u$ in the direction of the unit vector $(1/\sqrt 2,1/\sqrt 2)$, which can also be computed as $\nabla u(x,x)\cdot (1,1) = 4x$.