Let $f:\mathbb R^2\to\mathbb R$ be strongly convex in the first argument, and smooth in the second. Consider $$ \DeclareMathOperator*{\argmin}{argmin} x_*(y) = \argmin_x f(x, y)\,. $$ Can we somehow compute $\frac{dx_*}{dy}$ at a point $y$ without explicity computing the function $x_*$?
Whenever $f$ is also smooth in $x$, we have $$ 0=\frac{d}{dy}\frac{\partial f(x_*,y)}{\partial x} = \frac{\partial^2f}{\partial x^2}\frac{dx_*}{dy} + \frac{\partial^2 f}{\partial x\partial y} $$ and thus $$ \frac{dx_*}{dy} = -\left( \frac{\partial^2f}{\partial x^2} \right)^{-1} \frac{\partial^2 f}{\partial x\partial y} \,. $$ It would be nice to have something like this when $f$ doesn't have this additional regularity.
Just to provide an illustration: What would be a good approach for the following example?
$$ f(x,y) = |x-y|^2 + |x|\,. $$
The explicit solutions are $$ x_*(y) = \begin{cases} 0 & \text{if }y\in(-\frac12,\frac12) \\ y - \frac12 & \text{if }y\ge\frac12 \\ y + \frac12 & \text{if }y\le-\frac12 \end{cases}\,, \qquad \frac{d x_*}{dy} = \begin{cases} 0 & \text{if }y\in(-\frac12,\frac12) \\ +1 & \text{if }y>\frac12 \\ -1 & \text{if }y<-\frac12 \end{cases}\,. $$