Quick calculus problem! (Derivatives and tangent lines)

Find y' and the equation line to the graph at the indicated point.

e ^ xy - 2x = y + 1 Point: (0,0)


Find the equation(s) of the tangent line(s) at the point(s) on the graph of the equation.

y^3 - xy - x^3 = 2 where x = 1

These were just the harder problems in the packet that I wasn't able to get by myself. Please show all work. Thanks!

do dy/dx to both sides , y=y(x)

e ^ xy - 2x = y + 1
e^(xy) d/dx (xy) -2dx/dx = dy/dx +0

e^xy (xdy/dx +ydx/dx ) -2=dy/dx

e^xy (xdy/dx +y ) -2 =dy/dx

At (0,0)
1 ( 0 dy/dx +0) -2 = dy/dx
dy/dx= -2

2.- Similar .-
3y^2 dy/dx - (xdy/dx +y dx/dx) -3x^2 dx/dx = d/dx 2

3y^2 dy/dx - (xdy/dx +y ) -3x^2 = 0 , isolate dy/dx ,

dy/dx= (3x^2+y) / ( 3y^2-x)

At x=1 , isolate y from equation , y^3-1*y- 1^3 = 2

y^3-y-3=0

y~1.675 ( Numerical Method) , so the point is (1, 1.675 )
Plug in dy/dx = 4.675/ (7.4168)~ 0.63
The line is y-1.675 = 0.63 (x-1)

y~ 0.63 x +1.045