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\noindent{\bf Rudin, page 239/21b:}
We'll need the derivative:
$$
f'(x,y) = (6x^2 - 6x, 6y^2 + 6y) = 6(x(x-1),y(y+1))
$$
For functions $f:\R^2\to\R$, the implicit function theorem allows us to
solve $f(x,y) = 0$ locally for $x$ in terms of $y$ provided the partial 
derivative of $f$ with respect to $x$ does not vanish---i.e. provided 
$x\neq 0,1$.  Let us find out where on the zero level set of $f$ this
happens:
\begin{eqnarray*}
0 & = & f(0,y) = 2y^3 + 3y^2 \Rightarrow y = 0, - 3/2. \\
0 & = & f(1,y) = -1 + 2y^3 + 3y^2 \Rightarrow y= -1,1/2  
\end{eqnarray*}
So by the implicit function theorem, we can solve locally for $x$ in terms of 
$y$ except near the points $(0,0),(0,-3/2),(1,-1),(1,1/2)$.  

Now if the goal is to solve for $y$ in terms of $x$, then the implicit
function theorem allows us to do so at any point on the zero level set
where the partial derivative of $f$ with respect to $y$ vanishes.  The
problem points occur when $y=0$ or $y=-1$, and $2x^3 - 3x^2 + 2y^3 + 3y^2$.
More specifically we can solve for $y$ in terms of $x$ except at
$(0,3/2),(0,0),(1,-1),(-1/2,-1)$.

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