I just used WolframAlpha to derive a function $f(x)$ from the following equation: $$3x^3 =\Big(\frac{x+2y-1}{3}-x-y+1\Big)^2(1-x-2y)$$
I asked the program to isolate $y$, and I got an equation of the form $$y=f(x)=\sqrt[3]{p(x)}-\frac{q(x)}{\sqrt[3]{r(x)}}$$ or something along those lines. I'm not too concerned about what the actual result is, but I would like to know how to isolate a variable given an expression like the one above.
As said in comments, the equation reduces to a cubic in $y$, namely $$\left(\frac{31 x^3}{2}-6 x^2+6 x-2\right)+6\left(x-1\right)^2 y+\frac{9}{2} (x-1) y^2+y^3=0$$
If you follow the steps given here, you will find $$\Delta=-\frac{729}{2} x^3 \left(13 x^2-5 x+1\right)$$ which is of the same sign as $-x^3$. So, if $x>0$, $\Delta<0$ and the equation has one real root and two non-real complex conjugate roots. If $x=0$, the roots are $y=2$ (which is double root) and $y=\frac 12$. If $x<0$, then three real roots that you can express using $$p=-\frac{3}{4} (x-1)^2 \qquad \qquad q=\frac 14(53 x^3+3 x^2-3 x+1 )$$ and so on.