Dtyjlui

In the George F Simmons Calculus with Analytical Geometry, textbook, it is mentioned that the Partial derivatives of a function cannot be treated as fractions as in case of single variable function.

In the example provided it has been proved that,

from
$pV = nRT$

where,
- p - pressure

- T - Temperature

- V - Volume

- n - number of moles

- R - Universal gas constant

$dfrac{partial{p}}{partial{T}} dfrac{partial{T}}{partial{V}} dfrac{partial{V}}{partial{p}} = -1$

which is fine, but in the explanation, it is given as,
as the right hand side is -1 and not 1 we cannot treat the LHS of the above statement as fractions

If I am not wrong the 1st statement would mean, we cannot expand a partial differential as say,
$dfrac{d y}{d t} = dfrac{dy}{dx}dfrac{dx}{dt}$, where, $y=y(x)$ and $x = x(t)$,for a single variable function and we cannot represent the same for partial differentials the same manner if $y = y(x,t)$

What what is the significance of +1 or -1 for a differential to represented as a fraction?

Yes. $$dfrac{partial{x}}{partial{y}} dfrac{partial{y}}{partial{x}} neq dfrac{partial{x}}{partial{z}}$$
The equation above is an example to this phenomenon.

Although there's nothing wrong with the answer from mouthetics I would like to expand on something: ultimately what allows you to go from the lhs to rhs is the chain rule. For example, if $y = y(x, t)$ and $x = x(t)$ then

$$
frac{{rm d}y}{{rm d} t} = frac{partial y}{partial x}frac{{rm d}x}{{rm d}t} + frac{partial y}{partial t}
$$

However, consider this other case. Imagine a function $f(x, y, z) = 0$. Now suppose you can invert this relation to obtain $z = z(x, y)$, if you apply the chain rule you will get

$$
{rm d}z = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x{rm d}y tag{1}
$$

where I have used the script to indicate that variable remains fixed. Suppose now that you move along a path on the surface $f(x,y,z) = 0$, where $z$ is kept constant. In this case you can also invert the function $f$ to obtain $y = y(x)$ and

$$
{rm d}y = left(frac{partial y}{partial x}right)_z{rm d}x tag{2}
$$

Here it is more obvious why this notation is useful, it emphasizes the fact that we keep $z$ constant. Replace that in (1) with ${rm d}z = 0$ (remember, $z$ is constant) and you will get

$$
0 = left(frac{partial z}{partial x}right)_{y}{rm d}x + left(frac{partial z}{partial y}right)_x left(frac{partial y}{partial x}right)_z{rm d}x tag{3}
$$

And from here is trivial to obtain

$$
left(frac{partial x}{partial y}right)_z left(frac{partial y}{partial z}right)_x left(frac{partial z}{partial x}right)_y = -1 tag{4}
$$

Note that this is a result of consistently using the chain rule, and keeping track what function depends on what variables.

In your case just set $x = P, y = V, z = T$, the function $f(x,y,z)$ is called a equation of state