Does there exist an $x$ such that $3^x = x^2$? I tried solving for $x$ by using $x \log(3) = \log(x^2)
$$\log(3) = \frac{\log(x^2)}{x}$$
I'm stuck on this part. how do I isolate $x$ by itself?
Any help would be appreciated.

As Robert Israel wrote, the two roots of the equation are $$x_{\pm}=-2 \dfrac{W(\pm\ln(3)/2)}{\ln(3)}$$ The solution $x_-$ can be discarded if $x$ is supposed to be real because, for this case, the argument of Lambert function is smaller that $-\frac{1}{e}$.
As showed by Michael T, there is a solution such that $-1<x<1$ (in fact subbing in $x=0$, we find that $3^0-0=1$ is positive and then the root is somewhere between $x=-1$ and $x=0$). By inspection, you coul find an evn shorter range for the solution.
As you were told, the solution cannot be expressed using basic functions and, if you do not access Lambert function, only numerical methods could be used. Newton iterative scheme is probably the simplest and, starting from a reasonable guess, this will be updated according to $$x+\frac{x^2-3^x}{3^x \log (3)-2 x}$$ So, being very lazy and starting the iterative process at $x=0$, the successive iterates will then be $-0.910239$, $-0.703169$, $-0.686137$, $-0.686027$ which is the solution. 
More generally, the solution of an equation such as $$a b^x=x^c$$ has a solution which write $$x=-\frac{c W\left(-\frac{a^{\frac{1}{c}} \log (b)}{c}\right)}{\log (b)}$$
