Question

Find where the slope of the curve is defined: x^2y-xy^2=4

Answer 1

You do the implcit differentation, then solve for y' and check where this is defined. 
In your case: Differentiate implicitly: 2xy + x²y' - y² - x*2yy' = 0 
Solve for y': y'(x²-2xy) +2xy - y² = 0 
y' = (2xy-y²) / (x²-2xy) 
Check where defined: y' is not defined if the denominator becomes zero, i.e. 
x² - 2xy = 0 x(x - 2y) = 0 
This has formal solutions x=0 and y=x/2. Now we check whether these values are possible for the initially given definition of y: 
0^2*y - 0*y^2 =? 4 0 =? 4 
This is impossible, hence the function is not defined for 0, and we can disregard this. 
x^2*(x/2) - x(x/2)^2 =? 4 x^3/2 - x^3/4 = 4 x^3/4 = 4 x^3=16 x^3 = 16 x = cubicroot(16) 
This is a possible value for y, so we have a point where y is defined, but not y'. 
The solution to all of it is hence D - { cubicroot(16) }, where D is the domain of y (which nobody has asked for in this example :-). 
(Actually, the check whether 0 is in D is superfluous: If you write as solution D - { 0, cubicroot(16) }, this is also correct - only it so happens that 0 is not in D, so the set difference cannot take it out of there ...). 
If someone asks for that D, you have to solve the definition for y and find that domain - I don't know of any [general] way to find the domain without solving for the explicit function).