I want to isolate $x_2$. Is this correct? Lets say I have $(x_1,x_2)$ = (2,4) and the utility function $v_1(x_1,x_2)=g(x_1^2x_2)$ $$v_1(2,4)=g(2^2\cdot4) = g(16)$$
$$g(x_1^2x_2) = g(16) \Leftrightarrow x_1x_2 = \frac{g(16)}{g} \Leftrightarrow x_2 = \frac{16}{x_1}$$
My friend insists I am wrong, but won't explain further.
Your friend is right. :)
What does it mean to divide by $g$? $g$ is some (mysterious) function, not a number. What you've written is meaningless, and you also lost your exponent on the $x_1^2$. What you're trying to do will work only if $g(u)=g(v) \iff u=v$. As you might remember, functions satisfying this are called one-to-one. Then you can say $$g(x_1^2x_2)=g(16) \iff x_1^2x_2 = 16 \iff x_2=\frac{16}{x_1^2}.$$ Is your function $g$ one-to-one?