<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 can also differentiate explicitely:
$$y=(-3a^2x-x^3)^{1/3}.$$
$$y'=\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-3a^2-3x^2).$$
$$y''=-\frac{2}{3^2}(-3a^2x-x^3)^{-5/3}\cdot (-3a^2-3x^2)^2+\frac13 (-3a^2x-x^3)^{-2/3}\cdot (-6x)=$$
$$-\frac{2(a^2+x^2)^2}{y^5}-\frac{2x}{y^2}=\frac{-2(a^2+x^2)^2-2x(-3a^2x-x^3)}{y^5}=\frac{2a^2x^2-2a^4}{y^5}$$
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