How do you solve heavy equations in a organised and efficient way? I've made countless small mistakes on my exams that would hurt my grade. I have not been able to find a good guide on how to structure equation solving on paper so I thought I'd ask here. 
I've added a paper where I've solved for the possible max/min points in a multivariable equation. The paper is in my view very unorganized and I, therefore, wanted to ask you guys how you would solve it in a structural and organized way. Thanks in advance. 

 A: you have $$y(2x+y^2+y)=0$$ and $$x(x+3y^2+2y)=0$$ then we have the cases
$x=0$ or $y=0$
from $$2x+y^2+y=0$$ and $x+3y^2+2y=0$ we get
$$x=-3y^2-2y$$ and with this $$-6y^2-4y+y^2+y=0$$
A: My suggestion is to not forget that we are working with a system of two equations:
$$
\begin{cases}
f_x=0\\
f_y=0
\end{cases} \quad \iff \quad 
\begin{cases}
y(2x+y^2+y)=0\\
x(x+3y^2+2y)=0
\end{cases}
$$
so:
1) from the first equation we have $y=0$ that, subtituting in the second equation gives $x=0$ and we have found one solution of the system : $(x_1,y_1)=(0,0)$ 
2) from the second equation we have $x=0$ ad substituting in the first we find $y=-1$ or $y=0$. So we have another, different, solution of the system : $(x_2,y_2)=(0,-1)$ (note that, from here we have also the first solution $(x_1,y_1)$ that we just known).
3) if $x\ne0$ and $y \ne 0$ the system becomes:
$$\begin{cases}
2x+y^2+y=0\\
x+3y^2+2y=0
\end{cases}
$$
that, substituting $x$ from the second equation in the first, gives a new solution: $(x_3,y_3)=(\frac{3}{25},-\frac{3}{5})$
