Partial derivatives and the gradient

Partial derivative in $x$.
$$\frac{\partial f}{\partial x} = 2xy + y e^{xy}.$$

Partial derivative in $y$.
$$\frac{\partial f}{\partial y} = x^2 + x e^{xy}.$$

Gradient at $(1,0)$. With $x=1, y=0$:
$$\frac{\partial f}{\partial x}(1,0) = 0 + 0 \cdot e^0 = 0, \qquad \frac{\partial f}{\partial y}(1,0) = 1 + 1 \cdot e^0 = 2.$$
Hence
$$\nabla f(1,0) = (0, 2).$$

Remark. The gradient points in the direction of steepest increase of $f$ at the given point.