How to show that PDE is satisfied? 
How to solve this please? 
Attempt: 
$$a*\frac{\partial u}{\partial x} = \frac{\partial u}{\partial f} \frac{\partial f}{\partial x} = a(1)(\alpha) = a\alpha = (A) \\
b*\frac{\partial u}{\partial y} = \frac{\partial u}{\partial f} \frac{\partial f}{\partial y} = b(1)(\beta) = b\beta = (B)
$$
Therefore $(A)+(B) = aα + bß$ is the differential equation. I'm stuck from here (assuming I've done everything else correctly which I may not have). Thanks! I get very confused when performing partial differentiation where a function equals another function like this case.
 A: There are a couple things wrong.  To make things clearer, we should name the unnamed function that takes $(x,y)$ to $\alpha x+\beta y,$ so let $h(x,y) =\alpha x+\beta y.$  Then $u(x,y) = f(h(x,y))$ and the chain rules look like
$$\frac{\partial u}{\partial x} = \frac{df}{dh} \frac{\partial h}{\partial x} = \frac{df}{dh}\alpha$$
$$\frac{\partial u}{\partial y} = \frac{df}{dh} \frac{\partial h}{\partial y} = \frac{df}{dh}\beta$$
Note the plain $d$'s on the derivative of $f$.  Those are not partial derivatives.
Plug these into the PDE to get
$$a\frac{df}{dh}\alpha +b\frac{df}{dh}\beta = 0.$$
Now you just have to find "suitable constants" $\alpha$ and $\beta.$
If $\frac{df}{dh}=0$, then any values for $\alpha$ and $\beta$ will do.  If not, you can divide through by it and get 
$$a\alpha + b\beta = 0.$$
There are lots of solutions, but an easy one is $\alpha = b$ and $\beta = -a.$
A: $$
a\frac{\partial}{\partial x}f(\alpha x +\beta y)+b\frac{\partial}{\partial y}f(\alpha x +\beta y)\\
=(a \alpha+b \beta)f'(\alpha x +\beta y)=0 \\ 
\to \alpha=-{b \over a} \beta
$$
