Consider $$f(x,y) = \cases{ \text{sgn}(x) \dfrac{x^2-y^2}{x^2} & for $0 < |y| < |x|$ \cr 0 & otherwise }$$ Then $\displaystyle \int_{-1}^1 f(x,y)\ dy = 4 x/3$ for $-1 \le x \le 1$, so $\displaystyle \frac{\partial}{\partial x} \int_{-1}^1 f(x,y)\ dy = 4/3$, but $\dfrac{\partial}{\partial x} f(0,y) = 0$ so $\displaystyle \int_{-1}^1 \frac{\partial f}{\partial x}(0,y)\ dy = 0$.

Consider $$f(x,y) = \cases{ \text{sgn}(x) \dfrac{x^2-y^2}{x^2} & for $0 < |y| < |x|$ \cr 0 & otherwise }$$ Then $\displaystyle \int_{-1}^1 f(x,y)\ dy = 4 x/3$ for $-1 \le x \le 1$, so $\displaystyle \frac{\partial}{\partial x} \int_{-1}^1 f(x,y)\ dy = 4/3$, but $\dfrac{\partial f}{\partial x} (0,y) = 0$ so $\displaystyle \int_{-1}^1 \frac{\partial f}{\partial x}(0,y)\ dy = 0$.