Change of Variables - Writing the PDE in form of the new variables

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In a script, the author brings in the canonical form the PDE in $\mathbb{R}^2$ $$ \partial_{x x} u-2 \partial_{x y} u-3 \partial_{y y} u+\partial_y u=0 . $$ I only do not get this part:

(...) Therefore, the desired change of variables is $$ x^{\prime} =x, \qquad y^{\prime} =\frac{1}{2} x+\frac{1}{2} y $$ Under this change we have $$ \partial_{x x} u-2 \partial_{x y} u-3 \partial_{y y} u=\partial_{x^{\prime} x^{\prime}} u-\partial_{y^{\prime} y^{\prime}} u $$ and $$ \partial_y u=\frac{\partial x^{\prime}}{\partial y} \partial_{x^{\prime}} u+\frac{\partial y^{\prime}}{\partial y} \partial_{y^{\prime}} u=\frac{1}{2} \partial_{y^{\prime}} u . $$

How does he get the right hand sides of each part? Somehow the chain rule is used but I do not get the logic and the arithmetic behind the calculations. Thanks in advance.

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With the transformation $\xi=x,\eta=\frac{1}{2}x+\frac{1}{2}y$

we get the differential operators (using the chain rule)

$$\partial_x=\frac{\partial}{\partial{\xi}}\frac{\partial{\xi}}{\partial{x}}+\frac{\partial}{\partial{\eta}}\frac{\partial{\eta}}{\partial{x}}=\frac{\partial}{\partial{\xi}}+\frac{1}{2}\frac{\partial}{\partial{\eta}}$$

$$\partial_y=\frac{\partial}{\partial{\eta}}\frac{\partial{\eta}}{\partial{y}}=\frac{1}{2}\frac{\partial}{\partial{\eta}}$$

$$\partial{x x}=(\frac{\partial}{\partial{\xi}}+\frac{1}{2}\frac{\partial}{\partial{\eta}})^2=\frac{\partial^2}{\partial{\xi^2}}+\frac{\partial^2}{\partial{\xi}\partial{\eta}}+\frac{1}{4}\frac{\partial^2}{\partial{\eta^2}}$$

$$\partial{y y}=(\frac{1}{2}\frac{\partial}{\partial{\eta}})^2=\frac{1}{4}\frac{\partial^2}{\partial{\eta^2}}$$

$$\partial{x y}=(\frac{\partial}{\partial{\xi}}+\frac{1}{2}\frac{\partial}{\partial{\eta}})\cdot(\frac{1}{2}\frac{\partial}{\partial{\eta}})=\frac{1}{2}\frac{\partial^2}{\partial{\xi}\partial{\eta}}+\frac{1}{4}\frac{\partial^2}{\partial{\eta^2}}$$

Putting all together we finally get the transformed PDE $$(\partial{x x}-2\partial{x y}-3\partial{y y}+\partial{y})\cdot u(x,y)=(\partial{\xi \xi}-\partial{\eta \eta}+\frac{1}{2}\partial{\eta})\cdot u(\xi,\eta)=0$$

$$u_{\xi \xi}-u_{\eta \eta}+\frac{1}{2}u_{\eta}=0$$