Consider the ${3 \times 3}$ matrix $M$ whose entries consist of functions from $\mathbb{R}$ to $\mathbb{R}$ given by
$M_{ij} = \begin{cases} \cos^2\big(\frac{1}{2}\phi\big)+\sin^2\big(\frac{1}{2}\phi\big)\big(2(N_i)(N_i)-1\big), & \text{if $i=j$} \\ 2(N_i)(N_j)\sin^2\big(\frac{1}{2}\phi\big)+\epsilon_{ijk}(N_k)\sin(\phi), & \text{if $i \neq j$} \end{cases}$
where $\epsilon_{ijk}$ is the Levi-Cevita symbol with $\epsilon_{123}=1$ and $\phi$, $N_1$, $N_2$, and $N_3$ are class $C^3$ differentiable functions such that $N_1(x)^2+N_2(x)^2+N_3(x)^2=1$ for all $x \in \mathbb{R}$.
Let $t \in \mathbb{R}$ be fixed and denote by $Q \in M_{3 \times 3}(\mathbb{R})$ the matrix defined by evaluating each entry of $M$ at $t$. Also, denote by $Q',Q'' \in M_{3 \times 3}(\mathbb{R})$ the matrices defined by the entrywise first and second derivatives, respectively, of $M$ at $t$.
Now let $x,y,z \in M_{3 \times 1}(\mathbb{R})$ be any three column vectors. Consider the following matrix equation:
$Q''Q^Tx+2Q'y-Qz=0_{3 \times 1}$
How should I go about proving that the above equation implies $Q'=0_{3 \times 3}$ and, consequently, $z=0_{3 \times 1}$? Or does there exist a counterexample that I have yet to think of? It is probably important to note that $Q$ is a clockwise rotation matrix regardless of the value of $t$.
Your claim is false. Let me simplify a little by letting $$ X = Q^T x\\ Z = Qz $$ and noting that by choosing $x$ and $z$ appropriately, we can make $X$ and $Z$ have any value we want. So now your challenge is to show that in $$ Q''X + Q'y + Z = 0 $$ the only possibility is that $Q' = 0$ and $Z = 0$. But that's not true.
For instance, no matter what value $Q'$ has, we can pick $X = 0$, and choose any nonzero $y$, and then the equation becomes $$ Q' y + Z = 0 $$ which is easily solved by picking $Z = -Q' y$.
(To get $z$, we write $z = Q^{-1}Z = Q^t Z$; to get $x$, we write $x = QX$.)
Let me get even more explicit: Pick $N_1(x)= N_2(x) = 0$ and $N_3(x) = 1$ for all $x$; let $\phi(t) = t$. Then, letting $c$ and $s$ denote the since and cosine of $t$, we have something like $$ M = \pmatrix{c & s & 0 \\ -s & c & 0 \\ 0 & 0 & 1} $$ (where I might have the signs on the $s$ swapped). So $$ M' = \pmatrix{-s & c & 0 \\ -c & -s & 0 \\ 0 & 0 & 0} $$ and $$ M'' = \pmatrix{-c & -s & 0 \\ s & -c & 0 \\ 0 & 0 & 0} $$ Picking $t = \pi/2$, we have $s = 1$ and $c = 0$, so $$ Q = = \pmatrix{0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1}\\ Q' = \pmatrix{-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0}\\ Q'' = \pmatrix{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0} $$ Note in particular that $Q'$ is not zero .
Now pick $x = 0, y = \pmatrix{0 \\ 1 \\ 0};$ the equation you've highlighted then reads $$ \pmatrix{0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0} \pmatrix{0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1}^T \pmatrix{0\\0\\0} + 2 \pmatrix{-1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0} \pmatrix{0 \\ 1 \\ 0} - \pmatrix{0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1}\pmatrix{z_1 \\ z_2 \\ z_3} = \pmatrix{0 \\ 0 \\ 0}. $$
The first term is evidently zero; the second is $\pmatrix{0 \\ -2 \\ 0}$, and by picking $z$ to be \pmatrix{2 \ 0 \ 0}, we get equality, without $z$ being the zero vector.