Question

Given the point , find the slope of the curve or line that is tangent to the curve as requested.

y^6+x^3=y^2+11x, tangent at (0,1)

this is that I have so far

6y^5*y'+3x^2=2y+11

Answer 1

tiny but crucial mistake here
should be
\[6y^5y'+3x^2=2yy'+11\]

Answer 2

I ended up with something like
y' = -3x^2 +11 / (6y^5 -2y)