Do I need to substitute in the expressions before calculating this partial derivative? Find $\frac{\partial f}{\partial x}$ when: 
$f = 3xy^2z^3$, $y = 3x^2 + 2$, $z = \sqrt{x-1}$
I would need to replace $y$ and $z$ in $f$ before calculating the partial of $f$, correct? 
(The alternative would be find the partial of $y$ and $z$, square and cube them respectively and multiply all that by 3, which I think is incorrect, but I'm not sure)
 A: Yes, but it wouldn't be partial. $y$ and $z$ are both functions of $x$, so it doesn't really make sense to treat them as constants. So you would have $$f(x)=3x(3x^2+2)(x-1)^{3/2}=3(3x^3+2)(x-1)^{3/2}$$ which you can easily differentiate using the Product Rule.
A: We are free to either replace $y=y(x)$ and $z=z(x)$ or keep them in the calculation. If we decide to keep them in, which might also be convenient, we have to be aware that both $y$ and $z$ are functions of $x$ and have to treat them accordingly.
We consider
\begin{align*}
f(x,y,z)&=3xy^2z^3\\
y&=y(x)=3x^2+2\\
z&=z(x)=\sqrt{x-1}
\end{align*}
Differentiating $y=y(x)$ and $z=z(x)$ we obtain
\begin{align*}
y^\prime(x)=6x\qquad\qquad z^\prime(x)=\frac{1}{2\sqrt{x-1}}
\end{align*}

One way to do the partial differentiation is applying the product rule as following
  \begin{align*}
\color{blue}{\frac{\partial}{\partial x}f(x,y,z)}&=((3xy^2)z^3)^{\prime}\\
&=(3xy^2)^\prime z^3+3xy^2(z^3)^\prime\\
&=(3y^2+3x\cdot 2yy^\prime)z^3+3xy^2\cdot 3z^2z^\prime\\
&\,\,\color{blue}{=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime}\tag{1}\\
&=3(3x^2+2)^2(x-1)^{\frac{3}{2}}+6x(3x^2+2)6x(x-1)^{\frac{3}{2}}\\
&\qquad+9x(3x^2+2)^2\frac{1}{2}(x-1)^{\frac{1}{2}}\\
&\,\,\color{blue}{=\frac{3}{2}(x-1)^{\frac{1}{2}}(3x^2+2)(39x^3-30x^2+10x-4)}
\end{align*}

Similarly we obtain

\begin{align*}
\frac{\partial}{\partial x}f(x,y,z)&=(3x(y^2z^3))^{\prime}\\
&=(3x)^\prime y^2z^3+3x(y^2z^3)^\prime\\
&=3y^2z^3+3x(2yy^\prime z^3+3y^2z^2z^\prime)\\
&=3y^2z^3+6xyy^\prime z^3+9xy^2z^2z^\prime\\
\end{align*}
  giving the same result as in (1).

A: It is probably wanting you to use the Chain Rule for partial derivatives with intermediates.  This rule is, if $u = f(x, y)$, $x = g(s, t)$, and $y = h(s, t)$, then
$$ \frac{\partial u}{\partial s} = \frac{\partial u}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial s}$$
So, in your case, this would mean:
$$\frac{\partial f}{\partial x} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial x} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial x}$$
However, the $\frac{\partial f}{\partial x}$ is actually the total differential, $\frac{df}{dx}$.  $\frac{\partial f}{\partial x}$ refers to different things on each side of the equal sign.  On the left, it is the grand-total partial differential of $f$ with respect to $x$ (which, in this case, is the same as the total differential as well).  On the right, it is the partial differential of $f$ with respect to $x$ only as far as $x$ is concerned by itself.
