Very elementary question, $\frac{\mathrm{d}^2y}{\mathrm{d}x^2}$ if $x^2y^2+2\log y-3x+2=0$

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so as per title, this is embarrassingly elementary. Am writing out some answers for an undergrad economist taking a mathematical methods course. Most of the syllabus involves Simplex / Lagrangian optimisation / linear differences etc to give some idea of level. The above question came up, differentiating implicitly we find:

$$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y(3-2xy^2)}{2(1+x^2y^2)}$$

If we attack this again, we basically end up with a very ungainly formula owing to the multiplicity of terms generated by the products in the numerator and denominator (I'll spare your eyes seeing this, we're not talking quartic formula, but it ain't pretty!). While this yields a correct result, find it a bit surprising that an undergrad economics exam would have something this unwieldy / ugly in it, but maybe it does. Just wondering if there's something a bit more enlightened than brute force on this, a 'trick' of some kind, I suspect there probably isn't, but just in case someone has met this kind of thing in their travels. Thanks

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Deriving the implicit expression with respect to $x$ we have

$$x^2y^2+2\log y-3x+2=0\implies 2xy^2+2x^2yy'+2\frac{y'}{y}-3=0.$$

Deriving again the equality $2xy^2+2x^2yy'+2\frac{y'}{y}-3=0$ we get

$$ 2y^2+4xyy'+4xyy'+2x^2(y')^2+2x^2yy''+2\frac{yy''-(y')^2}{y^2}=0.$$

Finally get $y''$ from the above equality.