Verified Answer :

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\(\dfrac{\mathrm d}{\mathrm dx}\left[x^2y+3xy^3-x\right]=\dfrac{\mathrm d}{\mathrm dx}[3]\)
\(2xy+x^2\dfrac{\mathrm dy}{\mathrm dx}+3y^3+9xy^2\dfrac{\mathrm dy}{\mathrm dx}-1=0\)
\((x^2+9xy^2)\dfrac{\mathrm dy}{\mathrm dx}=1-2xy-3y^3\)
\(\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{1-2xy-3y^3}{x^2+9xy^2}\)

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