Suppose $f : \mathbb{R}^2 \to \mathbb{R}$ is twice differentiable with non-zero second partials. If for every $x \in \mathbb{R}$, $\exists$ unique $y^*(x)$ that solves $\underset{y \in \mathbb{R}}{\max} f(x,y)$, then show that $y^*(x)$ is increasing in $x$ if $\frac{\partial^2 f}{\partial x \partial y} > 0$.
My observations:
$y^*(x)$ is determined from $f_y(x,y) = 0$ (as $f_{yy}$ must be negative).
As you note, $y^*(x)$ satisfies the first and second order conditions (note the assumption of non-zero second partials implies a strictly negative second partial derivative in $y$):
$$f_y(x,y^*(x))=0,\quad (1)\\ f_{yy}(x,y^*(x))<0.\quad (2)$$
Implicit differentiation of $(1)$ gives
$${y^*}'(x)=-{\partial_x f_y(x,y)\over \partial_y f_y(x,y)}\bigg|_{y=y^*(x)}=-{f_{yx}(x,{y^*}(x))\over f_{yy}(x,y^*(x))}>0.$$