Essentially, you have worked out everything already, but there seems to be a bit of confusion about the definitions, so let me try to set this straight.
The differential of $f$ at the point $\mathbf{a} \in \mathbb{R}^2$ is the row matrix $ d_{\mathbf{a}}f = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}.$
Now if you write $d_{\mathbf{a}}f (\mathbf{u})$ for $\mathbf{u} = \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} \in \mathbb{R}^2$ you're meaning the matrix product $d_{\mathbf{a}}f (\mathbf{u}) = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) & \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 .$
On the other hand, $\nabla f (\mathbf{a})$ is the column vector $ \nabla f (\mathbf{a}) = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix}$ and when you're writing $\nabla f (\mathbf{a}) \cdot \mathbf{u}$ you're meaning the scalar product $\nabla f( \mathbf{a}) \cdot u = \begin{pmatrix} \frac{\partial}{\partial x} f(\mathbf{a}) \\\ \frac{\partial}{\partial y}f (\mathbf{a}) \end{pmatrix} \cdot \begin{pmatrix} u_1 \\\ u_2 \end{pmatrix} = \frac{\partial}{\partial x} f(\mathbf{a}) \cdot u_1 + \frac{\partial}{\partial y}f (\mathbf{a}) \cdot u_2 . $
So we see that for $f(x,y) = xy$ $d_{\mathbf{a}}f = \begin{pmatrix} y & x \end{pmatrix} \qquad \text{while} \qquad \nabla f (\mathbf{a}) = \begin{pmatrix} y \\\ x \end{pmatrix}.$
Now the confused reaction was due to the fact that the notation used here for the derivative of $f$ at the point $\mathbf{a}$ is often used as the directional derivative, and as you rightly pointed out in a comment, we have the relations $ D_{\mathbf{u}} f (\mathbf{a}) : = d_{\mathbf{a}} f (\mathbf{u}) = \nabla f(\mathbf{a}) \cdot \mathbf{u},$ and everything should be fine now, no?
Since you made the computations yourself already, I'll not repeat them here.