If you are searching a function
$$
T:\mathbb{R}^2 \to \mathbb{R}^2 \qquad T(x,y)=(T_1(x,y),T_2(x,y))
$$
such that
$$
dT = \begin{bmatrix} \frac{\partial T_1}{\partial x} & \frac{\partial T_1}{\partial y} \\ \frac{\partial T_2}{\partial x} & \frac{\partial T_2}{\partial y}\end{bmatrix}=
\begin{bmatrix} y & x \\ xy & x+y\end{bmatrix}
$$
than note that from the second row, we have:
$$
T_2(x)=\int xy dx =\frac{1}{2}x^2y+C(y)
$$
but
$$
\frac{\partial }{\partial y} \left(\frac{1}{2}x^2y+C(y) \right)
$$
cannot be $x+y$
Hint:
If you are searching a function $$ T:\mathbb{R}^2 \to \mathbb{R}^2 \qquad T(x,y)=(T_1(x,y),T_2(x,y)) $$ such that $$ dT = \begin{bmatrix} \frac{\partial T_1}{\partial x} & \frac{\partial T_1}{\partial y} \\ \frac{\partial T_2}{\partial x} & \frac{\partial T_2}{\partial y}\end{bmatrix}= \begin{bmatrix} y & x \\ xy & x+y\end{bmatrix} $$
than note that from the second row, we have: $$ T_2(x)=\int xy dx =\frac{1}{2}x^2y+C(y) $$ but $$ \frac{\partial }{\partial y} \left(\frac{1}{2}x^2y+C(y) \right) $$ cannot be $x+y$