The formulas for [latex]\partial w/\partial u[/latex] and [latex]\partial w/\partial v[/latex] are

[latex]\frac{\partial w}{\partial u}=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial u}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial u}[/latex]

[latex]\frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial v}.[/latex]

Therefore, there are nine different partial derivatives that need to be calculated and substituted. We need to calculate each of them:

[latex]\hspace{6cm}\begin{alignat}{2} \frac{\partial w}{\partial x} &= 6x -2y &\quad &\quad \frac{\partial w}{\partial y} &=-2x &\quad &\quad\frac{\partial w}{\partial z} &= 8z \\ \frac{\partial x}{\partial u} &= e^u\sin v &\quad &\quad \frac{\partial y}{\partial u} &=e^u\cos v &\quad &\quad\frac{\partial z}{\partial u} &= e^u \\ \frac{\partial x}{\partial v} &= e^u\cos v &\quad &\quad\frac{\partial y}{\partial v} &=-e^u\sin v &\quad &\quad \frac{\partial z}{\partial v} &= 0. \end{alignat}[/latex]

Now, we substitute each of them into the first formula to calculate [latex]\partial w/\partial u[/latex]:

[latex]\hspace{5cm} \begin{align} \frac{\partial w}{\partial u}&=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial u}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial u}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial u} \\ &=(6x-2y)e^u\sin v-2xe^u\cos v+8ze^u, \end{align}[/latex]

then substitute [latex]x(u,v)=e^u\sin v,y(u,v)=e^u\cos v[/latex], and [latex]z(u,v)=e^u[/latex], into this equation:

[latex]\hspace{5cm}\begin{align} \frac{\partial w}{\partial u}&=(6x-2y)e^u\sin v-2xe^u\cos v+8ze^u \\ &=(6e^u\sin v-2e^u\cos v)e^u\sin v-2(e^u\sin v)e^u\cos v+8e^{2u} \\ &=6e^{2u}\sin^2{v}-4e^{2u}\sin v\cos v+8e^{2u} \\ &=2e^{2u}(3\sin^2{v}-2\sin v\cos v+4). \end{align}[/latex]

Next, we calculate [latex]\partial w/\partial v[/latex]:

[latex]\hspace{5cm} \begin{align} \frac{\partial w}{\partial v}&=\frac{\partial w}{\partial x}\cdot \frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\cdot \frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\cdot \frac{\partial z}{\partial v} \\ &=(6x-2y)e^u\cos v-2x(-e^u\sin v)+8z(0), \end{align}[/latex]

then we substitute [latex]x(u,v)=e^u\sin v,y(u,v)=e^u\cos v,[/latex] and [latex]z(u,v)=e^u[/latex] into this equation:

[latex]\hspace{5cm}\begin{align} \frac{\partial w}{\partial v}&=(6x-2y)e^u\cos v-2x(-e^u\sin v) \\ &=(6e^u\sin v-2e^u\cos v)e^u\cos v+2(e^u\sin v)(e^u\sin) \\ &=2e^{2u}\sin^2{v}+6e^{2u}\sin v\cos v-2e^{2u}\cos^2{v} \\ &=2e^{2u}(\sin^2{v}+\sin v\cos v-\cos^2{v}). \end{align}[/latex]

Starting from the left, the function [latex]f[/latex] has three independent variables: [latex]x[/latex], [latex]y[/latex], and [latex]z[/latex]. Therefore, three branches must be emanating from the first node. Each of these three branches also has three branches, for each of the variables [latex]t[/latex], [latex]u[/latex], and [latex]v[/latex].

The three formulas are

[latex]\large{\frac{\partial w}{\partial t}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial t}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial t}}[/latex]

[latex]\large{\frac{\partial w}{\partial u}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial u}}[/latex]

[latex]\large{\frac{\partial w}{\partial v}=\frac{\partial w}{\partial x}\frac{\partial x}{\partial v}+\frac{\partial w}{\partial y}\frac{\partial y}{\partial v}+\frac{\partial w}{\partial z}\frac{\partial z}{\partial v}}.[/latex]