The curve defined by x^3+xy-y^2=10 has a vertical tangent line when x=?

Verteical is infinite rise/run or when the denomenator of dy/dx is 0

take derivitive, implicit differntation
3x^2+y+x(dy/dx)-2y(dy/dx)=0
3x^2+y=2y(dy/dx)-x(dy/dx)

find when denomenator is equal to 0

2y-x=0
2y=x
solve
subsitute 2y for x
(2y)^3+2y(y)-y^2=10
8y^3+2y^2-y^2=10
8y^3+y^2=10
at aprox y=1.03712
and
2y=x
2(1.03712)=x=2.07424

at about x=2.07424

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph