A nicer notion is that of the differential:
$$ \text{If} \qquad z = 5x + 3y \qquad \text{then} \qquad dz = 5\, dx + 3\,dy $$
Then if you decide to hold $y$ constant, that makes $dy = 0$, and you have $dz = 5 \, dx$.
Another notation that works well with function notation is that if we define
$$ f(x,y) = 5x + 3y$$
then $f_i$ means derivative of $f$ with respect to the $i$-th entry; that is
$$ f_1(x,y) = 5 \qquad \qquad f_2(x,y) = 3 $$
This doesn't work well with a common abuse of notation, though; sometimes people write $f(r,\theta)$ when they really mean "evaluate $f$ at the $(x,y)$ pair whose polar coordinates are $(r, \theta)$" rather than the 'correct' meaning of that expression "evaluate $f$ at $(r, \theta)$". So if you're in the habit of doing that, don't try to indicate derivatives by their position.
I confess I really dislike partial derivative notation; when one writes $\partial/\partial x$, one "secretly" means that they intend to hold $y$ constant, then when one passes it through the differential, one gets
$$ \frac{\partial z}{\partial x} = 5 \frac{\partial x}{\partial x} + 3 \frac{\partial y}{\partial x} = 5 \cdot 1 + 3 \cdot 0 = 5$$
However, the suggestive form of Leibniz notation starts becoming very misleading at this point; for example, let's compute other partial derivatives.
- $\partial z / \partial x = 5$, holding $y$ constant as the notation suggests
- $\partial x / \partial y = -3/5$, holding $z$ constant as the notation suggests
- $\partial y / \partial z = 1/3$, holding $x$ constant as the notation suggests
Then putting it together,
$$ \frac{\partial z}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z} = 5 \cdot \left(-\frac{3}{5}\right) \cdot \frac{1}{3} = -1 $$
This is a big surprise if you expect partial derivatives to behave similarly to fractions as their notation suggests!!!
Here's how you'd write everything precisely using the chain rule. There are actually two functions involved here. First is $f:\Bbb{R}^n\to\Bbb{R}$ and second is the map $\mu:\Bbb{R}\times\Bbb{R}^n\to\Bbb{R}^n$ defined as
\begin{align}
\mu(t,x)&:=tx.
\end{align}
So, $\mu$ is the "scalar-multiplication map". Now, we can consider the composite function $f\circ\mu:\Bbb{R}\times\Bbb{R}^n\to\Bbb{R}$ and ask what is its first partial derivative. The answer is given by the chain rule: for all $(t,x)\in\Bbb{R}\times\Bbb{R}^n$
\begin{align}
[\partial_1(f\circ \mu)](t,x)&=\sum_{j=1}^n(\partial_jf)(\mu(t,x))\cdot (\partial_1\mu^j)(t,x)\\
&=\sum_{j=1}^n(\partial_jf)(tx)\cdot x^j,
\end{align}
where the last line is simply because $\mu^j(t,x)=tx^j$.
Note that here, you shouldn't pay attention to the individual letters $t$ and $x$. They are just arbitrary curved symbols. Sure, it may be tradition to use these letters, but mathematically, one is not obligated to use them. I could just as well write: for all $(\ddot{\smile},@)\in\Bbb{R}\times\Bbb{R}^n$,
\begin{align}
[\partial_1(f\circ \mu)](\ddot{\smile},@)&=\sum_{\#=1}^n(\partial_{\#}f)(\ddot{\smile}@)\cdot @^j
\end{align}
As this crazy rewriting shows, logically speaking, one should not use notation such as $\partial_{x_j}f$ or $\frac{\partial f}{\partial x^j}$ in place of $\partial_jf$, simply because the symbol $x$ has no inherent mathematical meaning.
As for the remark about the FTC not being applicable, I'm not sure what your Professor meant, because the FTC is certainly applicable (assuming $f$ is $C^1$ for example): for all $x\in\Bbb{R}^n$,
\begin{align}
f(x)&=f(0)+\int_0^1\sum_{j=1}^n(\partial_jf)(tx)\cdot x^j\,dt
\end{align}
Best Answer
No, $\partial x$ cannot be understood as a product. If it could be, then you would get $$\frac{\partial y}{\partial x} = \frac{y}{x}$$ which obviously is not true in general.
The expression $\frac{\partial}{\partial x}$ is also known as derivation operator. This can be understood as follows:
Given a function of several variables, say $f(x,y)$, partial derivation with respect to $x$ can be seen as a map that maps the function $f$ to the function $\partial f/\partial x$. If you do so, you notice that this is a linear map, since the functions form a vector space, and $$\frac{\partial(\alpha f + \beta g)}{\partial x} = \alpha \frac{\partial f}{\partial x} + \beta \frac{\partial g}{\partial x}$$
Such linear functions on vector spaces are also known as operators. Now traditionally, the application of an operator to a vector is written like a product; this is because the prototype of this operation is multiplying a matrix with a vector.
So if you want to write down the derivation operator, you need a notation for it. And one common notation is to write the derivation operator as $\partial/\partial x$. The reason is probably exactly because it looks like application of a normal multiplication rule, so that $$\frac{\partial}{\partial x}f = \frac{\partial f}{\partial x}$$ and thus the rule is easy to remember and apply. On the other hand, as your question shows, it can easily mislead.
Another common notation for the partial derivative operator is $\partial_x$. This notation is usually preferred in the context of differential geometry. It has the advantage that you are not as easily misled about its meaning (and that is is less to write/type; for reasons that are outside the scope of this answer it's also a very natural notation in differential geometry), but has the slight disadvantage that you're less likely to figure out the meaning of $\partial_x f$ than of $(\partial/\partial x)f$ if you are not familiar with it.