You appear a little lost with your algebra there ... perhaps you are not seeing it because of all the x's where you are used to seeing a single constant?
Starting from: y^{2}-2y = x^{3}+2x^{2}+2x+3
You see that this is a quadratic in y ... solving it for y means finding the roots.
You've done this before.
Put it in standard form: ##ay^2+by+c=0##
then ##a=1, b=-2, c=x^{3}+2x^{2}+2x+3## and use the quadratic equation:
$$y\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$... and you get:
$$y\in\frac{2\pm\sqrt{4+4(x^{3}+2x^{2}+2x+3)}}{2}$$ ... and simplify.
In your case, the RHS is a
nice quadratic in y. So there's a shortcut.
##y^2-2y=(y-1)^2-1## so we can rewrite the original equation: $$(y-1)^2-1= x^{3}+2x^{2}+2x+3\\ \Rightarrow (y-1)^2=x^{3}+2x^{2}+2x+4$$ ... take the square-root of both sides and follow your nose.
This process is called "completing the square". Not all quadratics are nice like this.
There is also a cubic equation for if y appears to the third power.
http://en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots
... you could also try completing the cube. i.e. if you can express it as $$(y+\alpha)^3+\beta = f(x)$$ then you can just as $$y=(f(x)-\beta)^{1/3}-\alpha$$
More generally you may have ##\sum a_ny^n = f(x)## ... in which case you can get y in terms of f(x) by a combination of guessing and long division - like you learned to find the roots of a polynomial.
Really generally, a separable equation could end up as f(y)=g(x) ... i.e. the LHS is a function of y alone and the RHS is a function of x alone. How you approach this depends on f and g. There is no general approach.
