We have the following PDE:
$\frac{\partial W}{\partial \tau}=0.5\sigma^2\frac{\partial^2 W}{\partial x^2}$
We are going to look at a special solution of the from:
$W(\tau,x)=\tau^af((x-c)/\tau^b)$ where c is an arbitrary constant.
Substituting the solution we get:
$\tau^{a-1}(af-b\mu\frac{\partial f}{\partial \mu})=0.5\sigma^2\tau^{a-2b}\frac{\partial^2 f}{\partial \mu^2}$
where $\mu=(x-c)/\tau^b$
Now the part that I don't understand is why we can have a solution only if
$a-1=a-2b$
I think it is because you can clear in the following way: \begin{align*} \frac{af-b\mu\frac{\partial f}{\partial \mu}}{\frac{\partial^2 f}{\partial \mu^2}} = \frac{0.5\sigma^2\tau^{a-2b}}{\tau^{a-1}}. \end{align*}
Then the only way that the equality is fulfilled is that they are equal to a constant (since the left only depends on $\mu$, and the right side only depends on $\tau$), then: \begin{align*} \frac{af-b\mu\frac{\partial f}{\partial \mu}}{\frac{\partial^2 f}{\partial \mu^2}} = \frac{0.5\sigma^2\tau^{a-2b}}{\tau^{a-1}} =\lambda\in\mathbb{R}\Rightarrow \tau^{2b-1}=\frac{\lambda}{0.5\sigma^2}. \end{align*} But this would say that the variable tao is a specific value, and so $W$ is a function of one variable, which is a contradiction.