If $V$ is your pre-Hilbert space (which I am assuming is real), then let $$f:\ V \times V \to \Bbb R\ :\ x \mapsto \langle x, x \rangle$$ Then for each $x \in V, Df|_x$ is a linear map from $V \times V \to \Bbb R\ :\ v \mapsto 2\langle x, v\rangle$. Let $$g\ :\ \Bbb R \to \Bbb R\ :\ t \mapsto \sqrt t$$ Then $Dg|_t$ is a linear map from $\Bbb R \to \Bbb R\ :\ q \mapsto \frac{q}{2\sqrt t}$.

Hence $$\begin{align} D(g\circ f)|_x(v)&= Dg|_{f(x)}\circ Df|_x(v) \\&=Dg|_{\langle x, x\rangle}(2\langle x, v\rangle) \\&=\frac{2\langle x, v\rangle}{2\sqrt{\langle x, x\rangle}} \\&=\frac{\langle x, v\rangle}{\|x\|}\end{align}$$

You just have to remember that the differential is always a linear map. Even when we choose to represent it as a scalar, that is only because linear maps $\Bbb R \to \Bbb R$ are just scalar multiplication.

If $V$ is your pre-Hilbert space (which I am assuming is real), then let $$f:\ V \to \Bbb R\ :\ x \mapsto \langle x, x \rangle$$ Then for each $x \in V, Df|_x$ is a linear map from $V \to \Bbb R\ :\ v \mapsto 2\langle x, v\rangle$. Let $$g\ :\ \Bbb R \to \Bbb R\ :\ t \mapsto \sqrt t$$ Then $Dg|_t$ is a linear map from $\Bbb R \to \Bbb R\ :\ q \mapsto \frac{q}{2\sqrt t}$.

Hence $$\begin{align} D(g\circ f)|_x(v)&= Dg|_{f(x)}\circ Df|_x(v) \\&=Dg|_{\langle x, x\rangle}(2\langle x, v\rangle) \\&=\frac{2\langle x, v\rangle}{2\sqrt{\langle x, x\rangle}} \\&=\frac{\langle x, v\rangle}{\|x\|}\end{align}$$

You have to remember that the differential is always a linear map. Even when we choose to represent it as a scalar, this is only because linear maps $\Bbb R \to \Bbb R$ are just scalar multiplication.