$$dx=\dfrac{dy}{2}=\dfrac{du}{(-2x+y)u+2x^2+3xy-2y^2}$$
$$\dfrac{dx}{dy}=\dfrac{1}{2} \implies x=\dfrac{y}{2}+A$$
By sagemath software,
$$\dfrac{du}{dy}=\dfrac{(-2x+y)u+2x^2+3xy-2y^2}{2}=1.0 \, A^{2} - 1.0 \, A u + 2.5 \, A y$$
This is a first order linear inhomogeneous ODE.
$$u'-Au=A^2+2.5Ay$$
The integrating factor is $e^{-Ay}.$
The equation becomes $$e^{-Ay}u'-e^{-Ay}Au=e^{-Ay}(A^2+2.5Ay)$$
It is guaranteed that $$\dfrac{\partial (e^{-Ay}u)}{\partial y}=e^{-Ay}(A^2+2.5Ay)$$
$$e^{-Ay}u=-A e^{\left(-A y\right)} - \frac{2.5 \, {\left(A y + 1\right)} e^{\left(-A y\right)}}{A}+B$$
$$B=e^{-Ay}u+A e^{\left(-A y\right)} + \frac{2.5 \, {\left(A y + 1\right)} e^{\left(-A y\right)}}{A}$$
$$f(x-y/2)=e^{-Ay}u+A e^{\left(-A y\right)} + \frac{2.5 \, {\left(A y + 1\right)} e^{\left(-A y\right)}}{A}$$
$$f(x-\dfrac{y}{2}) -A e^{\left(-A y\right)}-\frac{2.5 \, {\left(A y + 1\right)} e^{\left(-A y\right)}}{A}=ue^{-Ay}$$
$$u=\dfrac{f(x-\dfrac{y}{2})}{e^{-Ay}}-A-\dfrac{2.5(Ay+1)}{A}$$ but does not work by substitution into the original PDE.
You didn't rearrange the differential equation correctly. Since
$$\dfrac{du}{dy}=A^{2} - \, A u + \frac{5}{2} A y$$
then we have
$$\frac{du}{dy}\color{red}{+}Au=A^2+\frac{5}{2}Ay$$
and multiplying by the integrating factor $e^{Ay}$ we have
$$\frac{d}{dy}\left(ue^{Ay}\right)=A^2e^{Ay}+\frac{5}{2}Aye^{Ay}$$
$$ue^{Au}=B+Ae^{Ay}+\frac{5}{2}ye^{Ay}-\frac{5}{2A}e^{Ay}$$ $$u=Be^{-Ay}+A+\frac{5}{2}y-\frac{5}{2A}$$
which is what JJacquelin obtained.