A problem about multivariable calculus and systems of equations How can I solve this problem?

Show that the system of equations
$$
\left\{\begin{array}{rcrcrcrcl}
3x & + & y & - & z & + & u^{2} & = & 0
\\
x & - & y & + & 2z & + & u\phantom{^{2}} & = & 0
\\
2x & + & 2y & - & 3z & + & 2u\phantom{^{2}} & = & 0
\end{array}\right.
$$
can be solved for $x,y,u$ in terms of $z$; for $x,z,u$ in terms of $y$; for $y,z,u$ in terms of $x$; but not for $x,y,z$ in terms of $u$.

My attempt: Note that $$x-y+2z+u=0 \quad + \quad 2x+2y-3z+2u=0 \implies 3x+y-z+3u=0$$Now $$3x+y-z+3u=0 \quad - \quad  3x+y-z+u^{2}=0 \implies 3u-u^{2}=0 \iff u=0 \quad \vee \quad u=3. $$
But, I think cannot conclude with that informations that the system cannot be solve for $x,y,z$ in terms of $u$.
Now, I think this problem is an applications of the Implicit function theorem, but I don't know model this problem for that way. For example, we can see that $$f(x,y,z,u):=(3x+y-z+u^{2},x-y+2z+u,2x+2y-3z+2u)$$
 A: I think you are correct about the implicit function theorem, but I can't quite finish the problem.  Consider the Jacobian matrix of the function $f$ that you defined:
$$Df=\begin{pmatrix}
3&1&-1&2u\\
1&-1&2&1\\
2&2&-3&2\\
\end{pmatrix}$$
The implicit function theorem tell us us that if when we cross out one of the columns, the resulting $3\times3$ matrix is invertible, then it is possible to solve for the other three variables in terms of the variable corresponding to the column we crossed out, in a neighborhood of a solution to the system.
Since you have shown that any solution must have $u=0$ or $u=3$, we have to consider both these cases.  To be sure that we can solve for $y,z,u$ in terms of $x$ we must check that both $$\begin{pmatrix}
1&-1&0\\
-1&2&1\\
2&-3&2\\
\end{pmatrix}$$  and
$$\begin{pmatrix}
1&-1&6\\
-1&2&1\\
2&-3&2\\
\end{pmatrix}$$
are invertible.  (I haven't done this.)  There are two cases to check for each of $y$ and $z$ also.
I don't know how to do the last part.  If we cross out the fourth column, we are left with a singular matrix, but the converse of the implicit function theorem isn't true.  I know of a partial converse, but notice that that talks about a differentiable solution.  The most I know how to conclude is that the implicit function theorem doesn't guarantee we can solve in terms of $u$.
It's been a long time since I studied this stuff, so it's quite possible, or rather quite likely, that there's something simple I'm overlooking.
