Determine whether the given relation is an implicit solution to the given equation.

sin(y)+xy-x^3=2-----relation

y''=(6xy'+((y')^3)sin(y)-2(y')^2)/((3x^2)-y)-----equation

The book says yes it is a solution but for some reason I am not getting it. You have to use implicit differentiation(duh).

What I did, which should be the right steps, is get the second derivative and solve for it(go from y-y'-y'' and isolate y''). Maybe I'm missing something.

This is what I got:

y''=(2y'sin(y)-2y'+6x)/(cos(y)+x)

Using implicit differentiation, we should be able to find y' in terms of x and y:

\sin{(y)}+xy-x^3=2

D\left \{ \sin{(y)}+xy-x^3 \right \}=D\left \{ 2 \right \}

D\left \{ \sin{(y)} \right \} + D\left \{ xy \right \}-D\left \{ x^3 \right \}=D\left \{ 2 \right \}

{y}' \cos{(y)} + x{y}' + y -3x^2 =0

{y}' = \frac{3x^2-y}{\cos{(y)} + x}

And differentiating a second time to find y'':

D\left \{{y}' \right \}= D\left \{\frac{3x^2-y}{\cos{(y)} + x}\right \}

{y}''=\frac{(6x-y')(\cos{(y)}+x)-(3x^2-y)({y}'\sin{(y)}+1)}{(\cos{(y)}+x)^2}

Now we can go back to the equation for y' and rearrange a little:

\cos{(y)} + x= \frac{3x^2-y}{{y}'}

Plugging that back into the equation for y'', we get

{y}''=\frac{(6x-y') \cdot {y}'}{3x^2-y}-\frac{({y}'\sin{(y)}+1) \cdot ({y}')^2}{3x^2-y}=\frac{6x{y}'+({y}')^3\sin{(y)}-2({y}')^2}{3x^2-y}

So I agree with your book. It definitely IS an implicit solution. Did that all make sense?