Derivative of function with respect to multivariable function Let 
\begin{align*}
  V\colon \mathbb R & \to \mathbb R \\
  x & \longmapsto
  V(x).
\end{align*}
and 
\begin{align*}
  w\colon \mathbb R^3 & \to \mathbb R \\
  (a,b,c) & \longmapsto
  w(a,b,c).
\end{align*}
I thought that using the chain rule we have $\dfrac{dV(w)}{dw(a,b,c)} = \dfrac{dV}{da} \dfrac{\partial a}{\partial w} + \dfrac{dV}{db} \dfrac{\partial b}{\partial w}+ \dfrac{dV}{dc} \dfrac{\partial c}{\partial w}$?
Numerically, I am finding that $\dfrac{dV(w)}{dw(a,b,c)} = \dfrac{1}{3}\cdot\left(\dfrac{dV}{da} \dfrac{\partial a}{\partial w} + \dfrac{dV}{db} \dfrac{\partial b}{\partial w}+ \dfrac{dV}{dc} \dfrac{\partial c}{\partial w}\right)$. 
Which is correct? Why do I need multiply by $\dfrac{1}{3}$?
 A: Using the chain rule should be correct. I've recently used it in multiple Partial Differential Equations exams, with the formula you've given:
$$\dfrac{dV}{dw(a,b,c)} = \dfrac{dV}{da} \dfrac{\partial a}{\partial w} + \dfrac{dV}{db} \dfrac{\partial b}{\partial w}+ \dfrac{dV}{dc} \dfrac{\partial c}{\partial w}$$
Wikipedia also says this is the correct way to apply the chain rule in higher dimensions: https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions
I'd say check your numerical findings again, or even show it to us, to see where this might have gone wrong.
A: Let $f:A\longrightarrow B$. We denote the Jacobian of $f$ at $x\in A$ by $J_f(x)$.
The chain rule for the Jacobian is
$$J_{f\circ g}(x) =  J_f\big(g(x)\big)\cdot J_g(x).$$
Applying this to your case we have
$$J_{V\circ w}(a,b,c) =  J_V\big(w(a,b,c)\big)\cdot J_w(a,b,c).$$
Now, $V$ is a single variable function and hence $J_V\big(w(a,b,c)\big) =  V'\big(w(a,b,c)\big)$, where $V'$ is the usual derivative.
On the other hand, $w:\mathbb R^3\longrightarrow \mathbb R$, so $J_w(a,b,c)$ is just the transpose of $\nabla w(a,b,c)$.
Therefore:
$$J_{V\circ w}(a,b,c) =  V'\big(w(a,b,c)\big)\cdot 
\left(
\frac{\partial w}{\partial a}(a,b,c),
\frac{\partial w}{\partial b}(a,b,c),
\frac{\partial w}{\partial c}(a,b,c)
\right)$$
A: Consider the following functions:
$$f = f(x,y)$$
$$g = g(f)$$
It's clear that $g$ can be written in terms of $(x,y)$ once the form of $f$ and $g$ is known. Thus using the chain rule,
$$dg = \dfrac{\partial{g}}{\partial{x}} dx + \dfrac{\partial{g}}{\partial{y}} dy$$
Also we can write:
$$\dfrac{\partial{g}}{\partial{x}} = \dfrac{\partial{g}}{\partial{f}} \dfrac{\partial{f}}{\partial{x}}$$
$$\dfrac{\partial{g}}{\partial{y}} = \dfrac{\partial{g}}{\partial{f}} \dfrac{\partial{f}}{\partial{y}}$$
Hence:
$$dg = \dfrac{\partial{g}}{\partial{f}}(\dfrac{\partial{f}}{\partial{x}} dx + \dfrac{\partial{f}}{\partial{y}} dy)$$
Finally:
$$\dfrac{dg(f)}{df(x,y)} = \dfrac{\partial{g}}{\partial{f}}(\dfrac{\partial{f}}{\partial{x}} \dfrac{dx}{df} + \dfrac{\partial{f}}{\partial{y}} \dfrac{dy}{df})$$
