<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] \begin{align}
y^3 - xy^2 + \cos xy=& 2\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-\sin xy(y+x\frac{dy}{dx}) =& 0\\
3y^2 \frac{dy}{dx} - y^2 - 2yx\frac{dy}{dx}-y\sin xy-x\sin xy\frac{dy}{dx} =& 0\\
\frac{dy}{dx}(3y^2 - 2xy - x\sin xy) =&y^2+y\sin xy
\end{align}
Hence
$$\frac{dy}{dx} = \frac{y^2+y\sin xy}{3y^2 - 2xy - x\sin xy}$$
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