<s>[INST] Find the derivative of $y^3-xy^2+\cos xy=2$ Find the derivative of $y^3-xy^2+\cos xy=2$
My Attempt:
$$y^3-xy^2+\cos xy=2$$
$$\dfrac {d}{dx} [y^3-xy^2+\cos xy]=\dfrac {d}{dx} [2]$$
$$3y^2.\dfrac {dy}{dx} -[1.y^2+2xy\dfrac {dy}{dx}] + (-\sin xy) \dfrac {dy}{dx}(xy)=0$$
$$3y^2 \dfrac {dy}{dx} - y^2 - 2xy\dfrac {dy}{dx} - \sin xy (y+\dfrac {dy}{dx} x)=0$$
How do I proceed further?
 [/INST] So your last line is:
\begin{align}
3y^2\dfrac{dy}{dx}-y^2-2xy\dfrac{dy}{dx}-\left(y+\dfrac {dy}{dx}x\right)\sin (xy)&=0\\
\implies\left(3y^2-2xy-x\sin(xy)\right)\dfrac{dy}{dx}&=y^2+y\sin(xy)\\
\implies\dfrac{dy}{dx}&=\dfrac{y^2+y\sin(xy)}{3y^2-2xy-x\sin(xy)}
\end{align}
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