= Solution
Write $u(x,y)=U(\xi,\eta)$. The <chain rule> gives $\partial_x=\partial_\xi+\partial_\eta$ and $\partial_y=2\partial_\xi+3\partial_\eta$. Consequently the coefficients of $U_{\xi\xi}$ and $U_{\eta\eta}$ cancel, while that of $U_{\xi\eta}$ is $12-25+12=-1$. The transformed <partial differential equation> is $-U_{\xi\eta}=1$. Integrating once in each variable gives the most general twice differentiable solution,
$$
\boxed{u(x,y)=-(x+2y)(x+3y)+F(x+2y)+G(x+3y),}
$$
where $F,G$ are arbitrary twice differentiable functions. The <change of variables> is invertible because its <Jacobian determinant> is $3-2=1$.
On $y=0$, the value condition gives $F(x)+G(x)=x^2$. Differentiating it yields $F'(x)+G'(x)=2x$. The derivative condition gives $-5x+2F'(x)+3G'(x)=x$. Subtraction therefore gives $G'(x)=2x$ and $F'(x)=0$. The constants in $F$ and $G$ cancel in their sum, leaving
$$
\boxed{u(x,y)=(x+3y)^2-(x+2y)(x+3y)=xy+3y^2.}
$$
Indeed $u_{xx}=0$, $u_{xy}=1$, $u_{yy}=6$, so the original operator gives $-5+6=1$, and the two initial-line conditions hold.
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