It's Calculus. Implicit differentiation is used.
Find the point besides the origin where the graph of xy=x+y^2 has a vertical tangent
Starting with xy = x + y2, implicit differentiation gives
y + xy' = 1 + 2yy'
We need to solve for y':
xy' - 2yy' = 1 - y
(x - 2y)y' = 1 - y
y' = (1 - y) / (x - 2y)
The slope is vertical if y' is undefined, which happens if x - 2y = 0 so that x = 2y.
Substituting this into the original equation gives 2y2 = 2y + y2. We then solve for y
y2 - 2y = 0
y(y - 2) = 0
y = 0 or y = 2.
Thus the possible points are (0,0) and (4,2)
We must check that in the expression for y', the numerator is not 0 for each point. Indeed that is the case here. Thus both points work.