Is it true that $\nabla_{x} (\min_y f(x,y)) = \nabla_x f(x,y^*)$, where $y^* = \arg\min f(x,y)$.

Question

I am sorry if this problem is very foolish. Suppose $f \colon \mathbb{R}^m \times \mathbb{R}^n \to \mathbb{R}$ be any continuous and differentiable function. Let $\phi(\cdot) = \min_{y} f(\cdot, y)$, then does the following equation hold for any $x \in\mathbb{R}^m$? $$\nabla_{x} \phi(x,y) = \nabla_x f(x,y^*), \quad y^* = \arg\min_y f(x,y).$$ If it is true, how to prove it? Otherwise, what conditions can guarantee this property?