Partial derivative of $\sin(xz)$: Is this making sense and is it a proper use of notation? The exercise here is to calculate $\dfrac{\partial}{\partial x} \sin(xz)$

The usual way is just considering $z$ as a constant and deriving, so it would be $z\cos(xz)$.

However, I want to solve it this way:
Define $a=xz$, then $\sin(a) = \sin(xz)$

$\dfrac{\partial}{\partial x} \sin(a) = \dfrac{\partial}{\partial a} \sin(a) \dfrac{\partial}{\partial x} xz = z\cos(a) = z\cos(xz)$.

Is this second way correct? My friend said it is not making sense and not a proper use of notation. But I do not understant why...
(Just to not leave this unanswered)

As comments suggested, this is a fair use of chain rule and there is nothing wrong. The other given answers are talking about problems that doesn't exist.
Yes.. but you are trying to replicate the chain rule in derivatives with the problem being it's not applied correctly. You can not use /∂a for Sin(a) and /∂x for xz.

You'll still get the same answer if you correct your technique slightly - 

Here you go - 

The Chain Rule says:
derivative of f(g(x)) = f'(g(x))g'(x)  

where 

f(g) = sin(g)  

Considering a=xz
Taking Sine on both sides 
sin(a) = sin(xz)

g(x) = xz

∂/∂x sin(a)=∂/∂x (Sin (xz)) 

-- Applying Chain Rule in Derivatives

0 = Cos(xz).∂/∂x(xz) 

=zCos(xz)