The first example is from: Widder, David V., "Advanced Calculus", Dover Books on Mathematics
Prerequisites: single-variable calculus, specifically u-substitution
Example 1:
Find $$\frac{\partial }{\partial x}\frac{sin(xy)}{cos(x+y)}$$
Solution:
We will need the quotient rule: $$\left ( \frac{f}{g} \right )' = \frac{f(x)g'(x) - f'(x)g(x)}{[g(x)]^{2}}$$
Setting $$f(x) = sin(xy)$$ and $$g(x) = cos(x+y)$$ we will need to find the derivatives of each.
Using the rule:
$$\frac{\mathrm{d} }{\mathrm{d} x}sin(u(x)) = cos(u(x))\frac{\mathrm{d} u}{\mathrm{d} x}$$
$$f'(x) = cos(xy)\frac{\partial }{\partial x}(xy)$$
$$f'(x) = ycos(xy)$$
And conversely, setting
$$g(x) = cos(x+y)$$
Using the rule:
$$\frac{\mathrm{d} }{\mathrm{d} x}cos(u(x)) = -sin(u(x))\frac{\mathrm{d} u}{\mathrm{d} x}$$
$$g'(x) = -sin(x+y)\frac{\partial }{\partial x}(x+y)$$
$$g'(x) = -sin(x+y)(1+0)$$
$$g'(x) = -sin(x+y)$$
Putting the parts together:
$$\frac{\partial }{\partial x}\frac{sin(xy)}{cos(x+y)} = \frac{sin(xy)(-sin(x+y))-(ycos(xy)cos(x+y))}{[cos(x+y)]^{2}}$$
This is usually sufficient for most professors. It is assumed that if you can get this far, you can simplify.