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
