Derivative of the $f(x,y)=\min(x,y)$

$$ f(x, y) = \min(x,y) = \begin{cases} x & \text{if } x \le y \\ y & \text{if } x \gt y \end{cases} $$

The function isn't differentiable along $y = x$, but the partial derivatives are straightforward otherwise.

$$ \frac{\partial f(x, y)}{\partial x} = \begin{cases} 1 & \text{if } x \lt y \\ 0 & \text{if } x \gt y \end{cases} $$

$$ \frac{\partial f(x, y)}{\partial y} = \begin{cases} 0 & \text{if } x \lt y \\ 1 & \text{if } x \gt y \end{cases} $$

Here is a plot of the function to help you see the derivatives and why it's not differentiable along $y = x$:

If $(a,b)$ is below the line $x=y$, then the function has value $y$ on a neighborhood of $(a,b)$, so the partial derivatives are $$\begin{align*} \left.\frac{\partial f}{\partial x}\right|_{(x,y)=(a,b)}&=0\\ \left.\frac{\partial f}{\partial y}\right|_{(x,y)=(a,b)} &= 1. \end{align*}$$ Symmetrically, if $(a,b)$ is "above" the line $x=y$, then the function has value $x$ on a neighborhood of $(a,b)$, so the partial derivatives are: $$\begin{align*} \left.\frac{\partial f}{\partial x}\right|_{(x,y)=(a,b)}&=1\\ \left.\frac{\partial f}{\partial y}\right|_{(x,y)=(a,b)} &= 0. \end{align*}$$

If $(a,b)$ is on the line $x=y$, then the function has value $y$ as we approach along a constant $y$ direction from the right, and value $x$ if we approach along a constant $y$ direction on the left. So the partial with respect to $x$ is $1$ from the left and $0$ from the right, hence does not exist at $(a,b)$. Similarly for $y$.

So the function is differentiable away from the line $x=y$, with values as given above.