Check for $x,y,z$ and $w$ if the equation system
$$3x + y-z+w^2 = 0$$
$$x-y+2z+w=0$$
$$2x+2y-3z+2w=0$$
can be solved respectively by the other three variables.
I tried to simplify the equation system $(III+II-I) \Longrightarrow w=0 \land w=3$ for example but i am not really sure if thats the right way.
I also had in mind to maybe calculate the roots of the equation system and then calculate the Jacobian Matrix and then check if $a_{i,j} \neq 0$ for the roots to use the implicit function theorem. But i am really unsure so any tips are appreciated
The system can be written as
$$\begin{pmatrix}3&1&-1\\1&-1&2\\2&2&-3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-w^2\\-w\\-2w\end{pmatrix}$$
As you've shown, $w=0\lor w=3$. So let us consider each of the cases:
$w\neq0\land w\neq3 \implies$ There are no solutions for $x,y,z$.
$w=3\implies$ Reducing $\left(\begin{array}{ccc|c}3&1&-1&-9\\1&-1&2&-3\\2&2&-3&-6\end{array}\right)$ using row reduction gives $\left(\begin{array}{ccc|c}1&0&\frac{1}{4}&-3\\0&1&\frac{-7}{4}&0\\0&0&0&0\end{array}\right)$
$\implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}-3\\0\\0\end{pmatrix}+\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$
$w=0 \implies$ Reducing $\left(\begin{array}{ccc|c}3&1&-1&0\\1&-1&2&0\\2&2&-3&0\end{array}\right)$ using row reduction gives $\left(\begin{array}{ccc|c}1&0&\frac{1}{4}&0\\0&1&\frac{-7}{4}&0\\0&0&0&0\end{array}\right)$
$\implies \begin{pmatrix}x\\y\\z\end{pmatrix}=\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$
Note that I was lazy and used a calculator to reduce the augmented matrix. If done algebraically, you'll get as the solution
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}f_1(w)\\f_2(w)\\f_3(w)\end{pmatrix}+\lambda\begin{pmatrix}\frac{1}{4}\\\frac{-7}{4}\\-1\end{pmatrix}\,\,\,,\,\,\,\lambda\in\mathbb R$$
in each case for some $f_1,f_2,f_3$.