For the equation $$PV=nRT$$ where $nR$ is constant
$$\frac{\partial P}{\partial V}\cdot\frac{\partial V}{\partial T}\cdot\frac{\partial T}{\partial P} = -1$$
If we were to simplify this, it implies $\frac{\partial P}{\partial P} = -1$
What does this mean? Change of $P$ with respect to $P$ is negative? What does that mean?
It doesn't simplify like that, because the different $\partial P$'s are not the same changes in $P$ and the same for the others. For example the first $\partial P$ is at constant $T$ while the second is at constant $V$. These generally don't change the same amount.
The more intuitive version of the triple product rule is
$$\frac{\partial z}{\partial y}=-\frac{\frac{\partial x}{\partial y}}{\frac{\partial x}{\partial z}}$$
or more explicitly
$$\left ( \frac{\partial z}{\partial y} \right )_x=-\frac{\left ( \frac{\partial x}{\partial y} \right )_z}{\left ( \frac{\partial x}{\partial z} \right )_y}.$$
The minus sign can be intuitively understood by considering the equation $dx=\frac{\partial x}{\partial y} dy + \frac{\partial x}{\partial z} dz = 0$, which can be rearranged to give the result provided that we interpret $dz/dy$ as $\frac{\partial z}{\partial y}$ (which is justified since we've assumed $dx=0$ in the first place).
For some physical intuition, think about the case $x=P,y=V,z=T$, in which case you find that the ratio on the right side is negative while the derivative on the left side must be positive.
Note that:
$$\frac{\partial P}{\partial V}=-\frac{nRT}{V^2}$$
$$\frac{\partial V}{\partial T}=\frac{nR}{P}$$
$$\frac{\partial T}{\partial P}=\frac{V}{nR}$$
thus:
$$\frac{\partial P}{\partial V}\cdot\frac{\partial V}{\partial T}\cdot\frac{\partial T}{\partial P} = -\frac{nRT}{V^2}\frac{nR}{P}\frac{V}{nR}=-\frac{nRT}{PV}=-1$$
Simplify $$\frac{\partial P}{\partial P} = -1$$
is meaningless.
