Since this is an equidimensional equation, determine all positive eigenvalues. \begin{align*} x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\ \phi(1) &= 0 \\ \phi(b) &= 0 \\ \end{align*} The given answer that I need to find the solution for is for $n = 1,2,\ldots$: \begin{align*} \lambda_n &= \left( \frac{n \pi}{\ln b} \right)^2 \\ \end{align*}
If we multiply by $1/x$ and rearrange we can get this in Sturm-Liouville form with $p(x) = x, q(x) = 0, \sigma(x) = 1/x$:
\begin{align*} x \frac{d^2\phi}{dx^2} + \frac{d\phi}{dx} + \frac{1}{x} \lambda \phi &= 0 \\ \frac{d}{dx} \left( x \frac{d\phi}{dx} \right) + \lambda \frac{1}{x} \phi &= 0 \\ \end{align*}
If we look at the Rayleigh quotient we can see there are no negative eigenvalues and we can manually verify that zero is not an eigenvalue.
Assume a trial solution of $\phi(x) = x^m, \phi'(x) = m x^{m-1}, \phi''(x) = m(m-1)x^{m-2}$ such that:
\begin{align*} x^2 \frac{d^2\phi}{dx^2} + x \frac{d\phi}{dx} + \lambda \phi &= 0 \\ m (m-1) x^m + m x^m + \lambda x^m &= 0 \\ \left( m^2 + \lambda \right) x^m &= 0 \\ \lambda &= -m^2 \\ \end{align*}
Since we know that $\lambda > 0$, then $m^2 < 0$, which means that $m = i r$ where $r \in \mathbb{R}$. However, this can't satisfy the boundary conditions:
\begin{align*} \phi(x) &= e^{ir} \\ \phi(1) = 0 &\neq 1^{ir} = 1 \\ \end{align*}
I'm stuck on what else to try.
You have to take the change of variable $y=\ln x$ and the equation becomes $$ \frac{d^2\phi}{dy^2}+\lambda\phi=0 $$ that is a standard form having as a solution $$ \phi(y)=A\cos\sqrt{\lambda} y+B\sin\sqrt{\lambda} y. $$ I think now you can go on from here.