If $y^3 + 3a^2x + x^3 = 0,$ then prove that $y'' + \frac{2a^2x^2}{y^5} = 0$ I am comfortable with second derivatives but I am just unable to set all the variables up in such a format that I get the latter (the part which needs to be proven). A hint would be a lot of help. 
 A: You want to use "implicit differentiation" to start, here:
$$ y^3 + 3a^2x + x^3 = 0 $$
$$ 3y^2y' + 3a^2 + 3x^2 = 0 \implies y^2y' + a^2 + x^2 = 0 $$
$$ 6yy'^2 + 3y^2y'' + 6x = 0 \implies 2yy'^2 + y^2y'' + 2x = 0 $$
Then a bit of algebra gives:
$$ y'' + \frac{2yy'^2 + 2x}{y^2} = 0 \quad \text{or} \quad y = 0 $$
$$ y'' + \frac{2y\left(\frac{-a^2 - x^2}{y^2}\right)^2 + 2x}{y^2} = 0 \quad \text{or} \quad y = 0 $$
$$ y'' + \frac{2\left(-a^2 - x^2\right)^2 + 2\left(-3ax^2 - x^3\right)x}{y^5} = 0 \quad \text{or} \quad y = 0 $$
If you then simplify further, you should find you get the desired result.
A: Differentiate twice,
$$y^2y'+a^2+x^2=0$$
and
$$2yy'^2+y^2y''+2x=0.$$
Then multiplying by $y^3$ and substituting,
$$2y^4y'^2+y^5y''+2xy^3=2(a^2+x^2)^2+y^5y''-2x(3a^2x+x^3)=0.$$
This simplifies to
$$y^5y''+2a^2(a^2-x^2)=0,$$ and the claim seems wrong.
A: You can also differentiate explicitely:
$$y=(-3a^2x-x^3)^{1/3}.$$
$$y'=\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-3a^2-3x^2).$$
$$y''=-\frac{2}{3^2}(-3a^2x-x^3)^{-5/3}\cdot (-3a^2-3x^2)^2+\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-6x)=$$
$$-\frac{2(a^2+x^2)^2}{y^5}-\frac{2x}{y^2}=\frac{-2(a^2+x^2)^2-2x(-3a^2x-x^3)}{y^5}=\frac{2a^2x^2-2a^4}{y^5}$$
A: by differentiating with respect to $x$ we get
$$3y^2\cdot y'+3a^2+3x^2=0$$ (I) from here we get
$$y'=-\frac{a^2+x^2}{y^2}$$ and $$y\ne 0$$
differentiating again we have
$$2y\cdot (y')^2+y^2y''+2x=0$$ (II)
now you can plug in the equation for $y'$ in (II) note that if we differentiate $$y^3$$ we get $$3y^2\cdot y'$$
