Original problem: consider the function $f = f_{a,b,c}(u,v,w)$: $$ f_{a,b,c}(u,v,w) = (v + T)^3 + v T (v+ T) - u^2 T - v w^2, \quad u,v,w \in\mathbb{R}, $$ where $$ T = -a u -b v- c w, $$ and $a,b,c\in\mathbb{R}$ are three parameters. I want to show that for any (possible) critical point $(u_0,v_0,w_0)$ of $f$, that is, $$ \nabla_{(u,v,w)}f(u_0,v_0,w_0) =0. $$ We have $$\mbox{rank}\,\nabla^2 f(u_0,v_0,w_0)\ge 2,$$ here $\nabla^2 f$ is the Hessian of $f$. Therefore, the stationary equation is $\nabla_{(u,v,w)}f(u_0,v_0,w_0) =0$, which is exactly the following three equations I posted. My plan was to show that there is only one the critical point $(0,0,0)$, but it seems not... so the relation on the Hessian should also be taken into account, and suppose on the contrary that rank$\nabla^2 f(u_0,v_0,w_0)$=1, then all $2\times 2 $ submatrix are degenerate. We shall get , actually, $\ge$ 6 equations. Then a tedious computation may be involved to show that all these equations do not have common solutions.
Let $a,b,c\in\mathbb{R}$ be three parameters satisfying $$abc\ne 0,\quad \mbox{and} \quad 2a^2\ne b^2.$$
Consider the following equations of $(u,v,w)\in\mathbb{R}^3$:
$$ (-c) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2vw, $$
$$ (-b) \bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = T^2 +w^2 +2vT, $$
$$ (-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT, $$ where $$T= -au-bv-cw.$$
Surprisingly, computer (WolframAlpha) tells me that for any $(a,b,c)$ as above, the solution of these equations is
$$
(u,v,w)= (0,0,0).
$$
How can I prove this claim? I tried many times but fail...
I notice that these equations are equivalent to
$$ 3(v+T)^2-2vT-(u^2+v^2) = \frac{2vw}{-c} = \frac{T^2+w^2+2vT}{-b} = \frac{2uT}{-a}, $$ which gives that (if $u\ne 0$ ) $$ T = \frac{a}{c}\frac{vw}{u}. $$
Another observation is that, if $(x_0,y_0,z_0)\ne (0,0,0)$ is a solution, then for any $t\ne 0$, $(tx_0,ty_0,tz_0)$ is a solution.
Okay see I reached a point. $$ \frac{T^2 + w^2 + 2vT}{-b}=\frac{2uT}{-a}$$ or $$ T^2 + w^2 + 2T(v-\frac{ub}{a})=0$$ Applying the quadratic eqn formula w.r.t $T$, we get $$T=-(v-\frac{ub}{a}) ± \sqrt {(v-\frac{ub}{a}) ^2 - w^2}$$ Now putting the value of $T=\frac{avw}{cu}$, $$\frac{avw}{cu} + (v-\frac{ub}{a})=±\sqrt{(v-\frac{ub}{a})^2 - w^2}$$ Squaring both sides, and further simplification gives that:- $$ (\frac{avw}{cu})^2+ 2\frac{avw}{cu}(v-\frac{ub}{a})=-w^2...(i)$$ From the above equation, $w$ can be taken common and be proved as $w=0$. It's a possible way. This means that $T=0$ as well. Further values can be computed from here. If $T=0$, then $$(-a)\bigg(3(v+T)^2-2vT-(u^2+v^2)\bigg) = 2uT \implies 2v^2-u^2=0$$ Here, they can have any value such that $$u=±\sqrt{2}v$$ I've come upto this point. Here $u$ and $v$ does satisfy $u=v=0$, but other values are possible too
If $w≠0$:- (2nd edit)
From my previous equation $(i)$, we can also say
$$ (\frac{av}{cu})^2w+ 2\frac{av}{cu}(v-\frac{ub}{a})=-w$$ After rearranging and simplifying this, we get $$w=\frac{2cu(vub-av^2)}{(av)^2+(cu)^2}$$ So if $a=b=c$, $$w=\frac{2u(vu-v^2)}{v^2+u^2}$$ So it's independent of $a,b,c$. For example, in this case, if you take $u=1,v=2$, you get $w=\frac{-4}{5}$.