Find the tangent line to the curve 2y 5 + x 2 y 3 = 1 + yex . at the point (0, 1).

First differentiate with respect to x:-

 10y^4.dydx  + x^2.3y^2.dy/dx + y^3.2x =  ye^x + e^x. dy/dx

dy/dx   =      ye^x - 2x y^3
             
             10y^4 + 3y^2x^2 - e^x

substituting  the point(0,1) into the above:-

dy/dx  =     1 e^0 - 0
                =  1/9
               10 + 0 - 1  

So the slope of the tangent  = 1/9
To find the equation of the tangent line:-
y - 1 = (1/9)(x - 0)

y = 1/9x + 1

or this could be written as 9y = x + 9