[tex]x^3y^2+\sin(x\ln y)+e^{xy}=0[/tex]
Differentiate both sides, treating [tex]y[/tex] as a function of [tex]x[/tex]. Let's take it one term at a time.
Power, product and chain rules:
[tex]\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}[/tex]
[tex]=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}[/tex]
[tex]=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}[/tex]
Product and chain rules:
[tex]\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}[/tex]
[tex]=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)[/tex]
[tex]=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)[/tex]
[tex]=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}[/tex]
Product and chain rules:
[tex]\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}[/tex]
[tex]=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)[/tex]
[tex]=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)[/tex]
[tex]=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}[/tex]
The derivative of 0 is, of course, 0. So we have, upon differentiating everything,
[tex]3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
Isolate the derivative, and solve for it:
[tex]\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}[/tex]
(See comment below; all the 6s should be 2s)
We can simplify this a bit by multiplying the numerator and denominator by [tex]y[/tex] to get rid of that fraction in the denominator.
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}[/tex]
