Find the values of $n$ so that the function $v=r^n(3\cos^2\theta-1)$ satisfies the relation$$\dfrac{\partial}{\partial r}\bigg(r^2\dfrac{\partial v}{\partial r}\bigg)+\dfrac{1}{\sin\theta}\dfrac{\partial v}{\partial\theta}\bigg(\sin\theta\dfrac{\partial v}{\partial\theta}\bigg)=0$$
I got, $$\dfrac{\partial}{\partial r}\bigg(r^2\dfrac{\partial v}{\partial r}\bigg)=n(n+1)r^n(3\cos^2\theta-1)$$
Also,$$\dfrac{1}{\sin\theta}\dfrac{\partial v}{\partial\theta}\bigg(\sin\theta\dfrac{\partial v}{\partial\theta}\bigg)=36\sin^2\theta\cdot\cos^2\theta \cdot r^{2n}$$
Adding them together and equating it to zero gives,$$n(n+1)(3\cos^2\theta-1)+9\sin^22\theta\cdot r^n=0$$
I don't know how to get $n$ from three unknowns, please help.
As a user pointed out in the comments, there is no solution for the expression as written. But consider the similar expression $$\dfrac{\partial}{\partial r}\bigg(r^2\dfrac{\partial v}{\partial r}\bigg)+\dfrac{1}{\sin\theta}\dfrac{\partial}{\partial\theta}\bigg(\sin\theta\dfrac{\partial v}{\partial\theta}\bigg)=0$$ Then the final expression would read $$[n(n+1)-6](3\cos^2 \theta -1) = 0$$ which does have a solution of $n=2,-3$. Given how chain rule questions are usually written for intro multivariable classes, I strongly suspect this is a typo. Check with your instructor.