<s>[INST] Analytically minimize $x^2 + y^2$ constrained to $(x+1)^3 =- y^2$ I have the following problem:
$$\text{min} \ x^2 + y^2$$
$$s.t. \ (x+1)^3 =- y^2$$
What I did was substituting, so I got the function $f(x) = x^2  - (x+1)^3 $ but I don't know how to get analytically to the minimum. Graphing the original problem gives $x = -1$ and $y=0$ as the solution but I can't get there using my new function, I did the first derivative equal to $0$ but that has no real solutions. I did Lagrange method on the original problem but it was nonsense too. Thanks for the help.
 [/INST] Believe in Algebra
If $x+1>0$, then we get a contradiction: $y^2<0$.
This immediately implies, $x+1≤0\implies x≤-1$ must be.
Then, we can define the function $f(x)$, such that  $$f(x)=x^2-(x+1)^3,~x≤-1.$$
Method $-1$
$$\begin{align}x≤-1\implies &\begin{cases}x^2≥1\\ (x+1)^3≤0\end{cases}\\\\
\implies &\begin{cases}x^2≥1\\ -(x+1)^3≥0
\end{cases}\\\\ \implies &x^2-(x+1)^3≥1.\end{align}$$
Finally, we conclude that,
$$\begin{align}&\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~\text{at}~|x|=1,~x+1=0 \\ \\
\implies &\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~~~~~~~~~~\text{at}~x=-1\\\ \end{align}$$
This means,
$$\min \left\{x^2+y^2 \mid (x+1)^3=-y^2\right\}=1,\\ 
\text{at}~x=-1,~y=0.$$

Method $-2$
$$\begin{align}x^2-(x+1)^3&=(x+1)^2-2x-1-(x+1)^3\\&=(x+1)^2-2(x+1)+1-(x+1)^3\\
&=(x+1)^2-(x+1)^3-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\\ \end{align}$$

Method $-3\text{a}$
$$\begin{align}x^2-(x+1)^3&=x^2-1-(x+1)^3+1=\\
&=(x-1)(x+1)-(x+1)^3+1\\
&=(x+1)(x-1-x^2-2x-1)+1\\
&=\underbrace{-(x+1)}_{≥0}\underbrace{(x^2+x+2)}_{>0}+1≥1.\end{align}$$
Method $-3\text{b}$
$$\begin{align}x^2-(x+1)^3&=-(x+1)(x^2+x+2)+1\\
&=-(x+1)\left((x+1)^2-(x+1)+2)\right)+1\\
&=-(x+1)^3+(x+1)^2-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\end{align}$$
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