Hang on.
I did it again, this time differentially and got 4 = 0. This does not explain why they don't intersect though?
It actually does - this is a classic case of proof by contradiction. The way it works is:
1. Assume some fact that you want to disprove (so in this case, we assume that the two graphs DO intersect)
2. Do some maths until you arrive at something that makes absolutely no sense
3. If the maths if you done in step 2 is all correct, then the reason for that must be that your assumption in step 1 is wrong
4. You're done. You proved it. Walk away happy.
By inserting y=x+1 into your equation, and finishing with 4=0, you've arrived at a contradiction - meaning that the graphs y=x+1 and -x^2+xy^2=x^2y CANNOT be equal.
hmm...
when I solved it I got x =0 and y =1, and also tried it on CAS and got the same answer.
But on the desmos graph of the two equations, they do not intersect at (0,1), so I am not sure what is going on.
The reason for this is that technically -x^2+xy^2=x^2y has an infinite amount of points that satisfy the equation at x=0. Let's see if the point (0,2) is on the graph:
-(0)^2+(0)(2)^2=(0)^2(2)
0+0=0
Since the result is true, that point exists. Let's try (0, 1):
-(0)^2+(0)(1)^2=(0)^2(1)
0+0=0
Result is true. Now, let's try the point (0, c), where c is any real number:
-(0)^2+(0)(c)^2=(0)^2(c)
0 + 0 = 0
Result is true. Technically, the question is wrong in that sense