Let $H$ be a real-valued function with three variables, $u,v$ be two real-valued functions with one variable. If we're asked to compute $$\frac{d}{dt}H\bigl(t,u(t),v(t)\bigr)$$
The tree diagram may seem like this:
However, the left most branch directly connects to the node $H$, which is a case of tree diagram that is not easily seen in the introductory calculus books. In these books, the trees often appears more "regular". So how to compute the result by this unusual tree?
In a more complicated way, how about to compute the partial derivative w.r.t. $x$ of $F\bigl(~~g(x,~h(y)),~~\phi(x,y,z)~~\bigr)$?
(Btw, I come from the background of real-analysis, thus the way to compute chain rule that I accustomed to is to use the matrix version $[Df(g(\mathbf{x}))]=[Df(g(\mathbf{x}))][Dg(\mathbf{x})]$, which is easy to remember and clear to compute. However, I just found it often too 'slow' to figure out which term we need.. So I think I should go back to use the introductory calculus way...)
When in doubt, write it fully: $$ t\longmapsto\pmatrix{t\cr u(t)\cr v(t)}\longmapsto H\pmatrix{t\cr u(t)\cr v(t)} $$ Let be $F$ the "inner" function: $$H(t,u(t),v(t)) = H(F(t)) = (H\circ F)(t).$$ By the chain rule: $$D(H\circ F)(t) = DH(F(t))DG(t) = DH(F(t))G'(t).$$ $$ ((\partial_x H)(F(t)),(\partial_y H)(F(t)),(\partial_z H)(F(t))) \pmatrix{1\cr u'(t)\cr v'(t)} = \cdots $$ Edit:
$\partial_x(F(g(x,h(y)),\phi(x,y,z)))$, i.e., the derivative of $x\longmapsto F(g(x,h(y)),\phi(x,y,z))$: $$x\longmapsto \pmatrix{g(x,h(y))\cr\phi(x,y,z)}\longmapsto F\pmatrix{g(x,h(y))\cr\phi(x,y,z)} $$ $$D(F\circ J)(x) = DF(J(x))DJ(x) = DF(J(x))J'(x) =$$ $$((\partial_1 F)(g(x,h(y)),\phi(x,y,z)),(\partial_2 F)(g(x,h(y)),\phi(x,y,z))) \pmatrix{(\partial_x g)(x,h(y))\cr(\partial_x\phi)(x,y,z)} = \cdots $$