Successive differentiation BSc 2 
Question : if $y^3 + x^3 - 3axy = 0$, show that $$y'' = -\frac{(2a^3 xy)}{(y^2-ax)^3}$$

How I did
I differentiated the whole expression twice (as shown below)


*

*$3y^2y'+3x^2-3x^2-3ay-3axy'=0$

*$6yy''+6x-3yy'-3ay'-3axy''=0$
Then I kept $y''$ on the left and transferred everything to right and plugged in the values of $y'$ and solved
I get 
$$y''=\frac{(y')(3y+3a)-(6x)}{6y-3ax}$$
On solving this further after putting values of $y'$ I'm not getting the desired thing: $y''=-2a^3xy/(y^2-ax)^3$

I got $$3y^2y'+3x^2-3ay-3axy'=0$$ or
$$y'=\frac{ay-x^2}{y^2-ax}$$
so 
$$y''=\frac{(ay'-2x)(y^2-ax)-(ay-x^2)(2yy'-a)}{(y^2-ax)^2}$$
$$y''=\frac{y'(-a^2x-ay^2+2xy)-2xy^2+ax^2+a^2y}{(y^2-ax)^2}$$
$$y''=\frac{(ay-x^2)(-ax^2-ay^2+2x^2y)-(2xy^2-ax^2-a^2y)(y^2-ax)}{(y^2-ax)^3}$$
Expanding all i got
$$y''=\frac{-2a^3xy+6ax^2y^2-2x^4y-2xy^4}{(y^2-ax)^3}$$
