Is there a chain rule for partial derivatives?

Yes, there is.

Suppose you have three functions: #y(x,z)#, #x(t)#, and #z(t)#.

#y = x + z#
#x = 2t#
#z = t^2#

The function #y# varies according to #x# and #z#, while #x# and #z# are function themselves that vary with #t#. When you take the partial derivative with respect to #x#, for instance:

#(dely)/(delx) = 1*color(green)((delx)/(delx)) + 1*color(green)((delz)/(delx))#

But since #z# is not a function of #x#, nothing happens to it.

#= 1 + z#

So you still see the chain rule, but it may not be obvious, and here it looks a bit redundant. #(delx)/(delx)# merely becomes a #1# multiplier in the equation.

This is similar to the chain rule you see when doing related rates, for instance.

However, if you take the exact differential with respect to #t#, which is like a variable "indirectly associated" with #y# or "embedded" within #y#, it is easier to see the effects of the chain rule here, where you have embedded partial derivatives (inexact differentials) due to the fact you have multiple variables:

#(dy)/(dt) = d/(dt)[x + z]#

#= (dely)/(delx)cdotcolor(green)((delx)/(delt)) + (dely)/(delz)cdotcolor(green)((delz)/(delt))#

You may see that this is a convenient notation that allows you to see the rationale for formulating the way you take these partial derivatives:

#(dy)/(dt) = (dely)/cancel(delx)cdotcolor(green)(cancel(delx)/(delt)) + (dely)/cancel(delz)cdotcolor(green)(cancel(delz)/(delt))#

That aside, you get, from #(dely)/(delx) = 1# and #(dely)/(delz) = 1#:

#= 1*color(green)((delx)/(delt)) + 1*color(green)((delz)/(delt))#

#= color(blue)(2 + 2t)#