Let $f(x,y,t) :=u(r)\cos \omega t$, where $r= \sqrt{x^2 +y^2}$. Physics tells us the following: For $f(x,y,t)$ to describe a vibrating membrane, with $f(x,y,t)$ telling how high the mem- brane is above the rest position at point $(x,y)$ and at time $t,f$ must satisfy the condition $$\dfrac{\partial^2 f}{\partial x^2} + \dfrac{\partial^2 f}{\partial y^2} = c^{−2} \dfrac{\partial ^2 f}{\partial t^2}$$ with $c$ the wave propagation speed in the membrane.
Your job is simply to use the multivariable chain rule and obtain an ODE for $u$ from the partial differential equation for $f$. Compare the ODE so obtained with the Bessel ODE.
I think that I understand the multivariable chain rule, but I am confused about whether to take the partial derivative of f with respect to $x$ and $y$ or $r$...or if I even need to start from the original function of $f$ or work just from the pde condition given.
Just take the given pde for $f$ and plug in the form of $f$ that you know.
For example $\partial_{x} f = u'(r) (\partial_x r) \cos \omega t = u'(r) \frac{x}{r} \cos \omega t.$ Do the same to compute $\partial_{xx} f, \partial_{yy} f, \partial_{tt} f$ (the last one is easy since $r$ doesn't depend on $t$!). Once you have all these partial derivatives in terms of $u$, do you see how your original pde turns into an ode for $u$ just by plugging in the expressions in terms of $u$ for $\partial_{xx} f, \partial_{yy} f, \partial_{tt} f$?