$h(x)=\langle f(x),g(x)\rangle$ is differentiable in $c$ and $Dh(c)u=\langle Df(c)u,g(c)\rangle+\langle f(c),Dg(c)u\rangle$ with $u\in \mathbb{R}^m$.

Let $$G\subset\mathbb{R}^m$$ where $$G$$ is open and $$c\in G$$. If $$f,g:G\to\mathbb{R}^n$$ differentiable in $$c$$ then $$h:G\to\mathbb{R}$$ defined by $$h(x)=\langle f(x),g(x)\rangle$$ is differentiable in $$c$$ and $$Dh(c)u=\langle Df(c)u,g(c)\rangle+\langle f(c),Dg(c)u\rangle$$ with $$u\in \mathbb{R}^m$$.

I'm somewhat lost in how to proof this. I know that since $$f,g$$ are differentiable in $$c$$ then we have existence of $$Df(c)$$ and $$Dg(c)$$. As well since $$\langle\cdot ,\cdot\rangle$$ is bilinear , then $$D\langle u,v\rangle(x,y)= \langle x,v\rangle+\langle u,y\rangle$$.

I have tried connecting these to things but haven't been able to get anywhere.

HINT

$$\frac{|h(c+y)-h(c)- \langle Df(c)(y),g(c)\rangle - \langle f(c),Dg(c)(y)\rangle|}{\|y\|}$$ 1. Add and subtract $$\langle f(c+y),g(c)\rangle$$ $$\langle f(c+y),Dg(c)(y)\rangle$$ inside the absolute value.

2.Use the continuity of $$f$$ at $$c$$ and the differentiabiliy of $$f,g$$ at $$c$$ and the Cauchy Schwartz inequality $$|\langle x,y\rangle| \leq \|x\|\|y\|$$ and the triangle inequality and you will have your conclusion.

Note that $$\|f(c+y)\| \to^{y \to 0} \|f(c)\|$$

Let $$\omega(\cdot, \cdot) = \langle \cdot, \cdot \rangle$$ denote the inner product (for ease of typing). The idea is that you want to somehow be able to express $$h$$ as a composition of $$\omega,f,g$$. To do this properly, introduce the following "canonical injections": first define $$\iota_1 : \Bbb{R}^n \to \Bbb{R}^n \times \Bbb{R}^n$$ by the rule $$\iota_1(x) = (x,0)$$ and $$\iota_2 : \Bbb{R}^n \to \Bbb{R}^n \times \Bbb{R}^n$$ by the rule $$\iota_2(x) = (0,x)$$. Then, I leave it to you to verify that \begin{align} h &= \omega \circ(\iota_1 \circ f + \iota_2 \circ g) \tag{**} \end{align} (be sure to check that both sides have the same domain $$G$$ and same codomain $$\Bbb{R}$$, and that all the compositions actually make sense, and I hope you realise where the addition $$+$$ on the RHS is taking place etc).

Now, computing the derivative is really a direct application of the chain rule which says that for any differentiable maps $$\phi,\psi$$, and any point $$c$$ (in an appropriate domain) we have \begin{align} D(\phi \circ \psi)_c = D \phi_{\psi(c)} \circ D \psi_c \end{align} (I simply put the point of derivative in subscript to avoid too many brackets, because these are linear maps and hence can act on a vector). If we apply this to our formula for $$h$$, we see that for all $$c \in G$$, \begin{align} Dh_c &= D \omega_{(\iota_1 \circ f + \iota_2 \circ g)(c)} \circ \bigg( D(\iota_1)_{f(c)} \circ Df_c + D(\iota_2)_{g(c)} \circ Dg_c \bigg) \\ &= D \omega_{(f(c), g(c))} \circ \bigg( \iota_1 \circ Df_c + \iota_2 \circ Dg_c \bigg) \tag{*} \end{align} In $$(*)$$, I used the fact that $$\iota_1$$ is a (continuous) linear transformation, hence at every point, it is its own derivative; i.e for all $$a$$, $$D(\iota_1)_a = \iota_1$$, and likewise for $$\iota_2$$. So far, $$(*)$$ has been written down as an equality of linear transformations $$\Bbb{R}^m \to \Bbb{R}$$, but we can now evaluate both sides on an arbitrary vector $$u \in \Bbb{R}^m$$ to conclude that \begin{align} Dh_c(u) &= D \omega_{(f(c), g(c))} \bigg(\left(Df_c(u), 0 \right) + \left(0, Dg_c(u) \right) \bigg) \\ &= D \omega_{(f(c), g(c))} \bigg(\left(Df_c(u), Dg_c(u) \right) \bigg) \\ &= \omega \left(Df_c(u), g(c) \right) + \omega \left( f(c), Dg_u(c)\right) \end{align}

Btw even if all this composition business seems abstract, I would highly suggest you learn to deal with it, because it's one of those things which in the beginning may be pretty difficult to keep track of where each object is defined etc, but its really very helpful, and really, this is just a generalisation of the product rule from standard single variable calculus, which in turn is really just a consequence of the chain rule.

I could have merely written down the equation $$(**)$$, and said "apply the chain rule from here", because that's really all that's happening, but I merely fleshed out the details of how to apply the chain rule in this situation. Also, I suggest you take a look at this answer of mine, where I state the theorem in a more general context, and also show how to use this in an actual computation. Once you understand this idea, I invite you to state and prove a generalization of this to a "product" involving $$n$$ terms.