<s>[INST] 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. 
 [/INST] 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.
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