Find $ y''$ and $ y'''$ if $b^2x^2+a^2y^2=a^2b^2$ answers :- $$y''=-\frac{b^4}{a^2y^3}$$ and  $$ y'''=-\frac{3b^6x}{a^4y^5}$$i have tried solve in this manner:-$$b^2x^2+a^2y^2=a^2b^2  -1.)$$diff wrt x :$$2b^2x+2a^2y'=0$$ $$a^2y'=-b^2x$$$$\frac{dy}{dx}=\frac{-b^2x}{a^2y}$$on substituting eq -1 we get :-$$\frac{dy}{dx}=\frac{-b^2x}{a \sqrt{a^2b^2-b^2x^2}}$$$$\frac{dy}{dx}=-a^{-1}.b^2.x.(a^2b^2-b^2x^2)^{-\frac{1}{2}}$$on diff wrtx :- $$y''=\frac{-b}{a}[(a^2b^2-b^2x^2)^{-\frac{1}{2}}+{-\frac{1}{2}}(a^2b^2-b^2x^2)^{-\frac{3}{2}}.-2b^2.x.x]$$$$y''=\frac{-b}{a}[(a^2b^2-b^2x^2)^{-\frac{1}{2}}+(a^2b^2-b^2x^2)^{-\frac{3}{2}}.b^2.x^2]$$now substituting$(a^2b^2-b^2x^2)$as $a^2y^2$ from eq -1.):- $$y''=b^2\frac{(a^2y^2)^3+ax^2}{a(a^2y^2)^{\frac{3}{2}}}$$**This was all i could have done someone please help**

The question is not very clear but it seems that you are writing $y_1$ for the derivative of $y$ and $y_2$ for the second derivative.
If I am correct in this, then
$$\eqalign{b^2x^2+a^2y^2=a^2b^2\quad
  &\Rightarrow\quad 2b^2x+2a^2yy'=0\cr
  &\Rightarrow\quad 2b^2+2a^2(yy''+(y')^2)=0\qquad(*)\cr
  &\Rightarrow\quad y''=-\frac{b^2}{a^2y}-\frac{(a^2yy')^2}{a^4y^3}\cr
  &\Rightarrow\quad y''=-\frac{a^2b^2y^2+b^4x^2}{a^4y^3}\cr
  &\Rightarrow\quad y''=-\frac{b^2(a^2b^2)}{a^4y^3}=-\frac{b^4}{a^2y^3}\ .\cr}$$
You can use similar ideas to find $y'''$: it will probably be easiest to begin by differentiating $(*)$.
