Showing that a differential is nonzero 
Problem Statement: Let $g:U\rightarrow \mathbb{R}^{n}$ be a $C^{1}$ map ($U\subset \mathbb{R}^{n}$ open), with $dg(\mathbf{x})\in GL(\mathbb{R}^{n})$ for all $\mathbf{x}\in U$. Suppose $\mathbf{y}_{0}\notin g(U)$, and let $\psi(\mathbf{x})=\lvert g(\mathbf{x})-\mathbf{y}_{0}\rvert^{2}$. Show that $d\psi(\mathbf{x})\neq \mathbf{0}$, for any $\mathbf{x}\in U$.

I have been working through this problem (from Fleming) but I am kind of lost. First, I do not know if I am computing $d\psi(\mathbf{x})$ correctly. I thought by the chain rule, we have
$$d\psi(\mathbf{x})=2\lvert g(\mathbf{x})-\mathbf{y}_{0}\rvert\circ dg(\mathbf{x}).$$
Then we must show that for any $\mathbf{x}\in U$ we may find $\mathbf{v}\in \mathbb{R}^{n}$ so that $d\psi(\mathbf{x})[\mathbf{v}]$.
We know that $dg(\mathbf{x})[\mathbf{v}]\neq \mathbf{0}$ if $\mathbf{v}\neq \mathbf{0}$ since $dg(\mathbf{x})\in GL(\mathbb{R}^{n})$. But, (if I even computed $d\psi$ correctly) I am not sure where $2\lvert g(\mathbf{x})-\mathbf{y}_{0}\rvert$ comes into play, or my choice for $\mathbf{x}$. Since $\mathbf{y}_{0}\notin g(U)$, then $\lvert g(\mathbf{x})-\mathbf{y}_{0}\rvert\neq 0$, but that seems too trivial, because then I would have $d\psi(\mathbf{x})[\mathbf{v}]=2\lvert g(\mathbf{x})-\mathbf{y}_{0}\rvert\circ dg(\mathbf{x})[\mathbf{v}]\neq\mathbf{0}$ for any $\mathbf{x}$ and $\mathbf{v}\neq \mathbf{0}$.
I was given a hint that if $A\in GL(\mathbb{R}^{n})$ then $A^{T}[\mathbf{w}]\neq \mathbf{0}$ if $\mathbf{w}\neq \mathbf{0}$, and so $\mathbb{v}=A^{T}[\mathbf{w}]$ satisfies $\langle \mathbf{w},A[\mathbf{v}]\rangle\neq 0$. I assume I should be applying this fact to $dg(\mathbf{x})$, but I don't understand how the inner product is being used relative to this problem.
Any suggestions are appreciated!

It is sometimes better to write down the functions involved explicitly.
In this problem there are two functions, one is $g : U \to \mathbb{R}^n$ and the other is a function $f : U \to \mathbb{R}$ defined by $$f(u) = \lVert u - y_0 \rVert^2 = \sum_{i=1}^n (u^i - y_0^i)^2$$ where $u = (u^1,\ldots, u^n)$ and $y_0 = (y_0^1,\ldots,y_0^n)$. Then the derivative of $f$ is $$ df_u = 2(u^1-y_0^1,\ldots, u^n-y^n).$$ 
The function $\psi$ is just $\psi = f \circ g : U \to \mathbb{R}$. Applying chain rule we have $$d\psi_x = df_{g(x)}\circ dg_x $$ for all $x \in U$.
If there is $x \in U$ with $d\psi_x = df_{g(x)}\circ dg_x = 0$ we can conclude that $df_{g(x)} =0$ because $dg_x$, by hypothesis, is a non-singular linear map. Therefore, $$ df_{g(x)} = 2(g^1(x)-y_0^1,\ldots, g^n(x)-y^n_0)=0$$ which of course implies that $g(x) = (g^1(x),\ldots, g^n(x))=y_0$, a contradiction.
