What is the 3rd degree Taylor polynomial expanded at $x=0$ of the function $x\sin(y)=y+\sin(x)$? 
What is the 3rd degree Taylor polynomial expanded at $x=0$ of the function $x\sin(y)=y+\sin(x)$?

I am wondering if there is an easy way to do this?
 A: I would do it by the stupid-minded method of formally expanding the sine series out far enough to get an equation in series, but ignoring terms of higher degree than you’re interested in. Your given equation is:
$$
y+x-\frac{x^3}6+\cdots=x(y - \frac{y^3}6+\cdots)=xy-\frac{xy^3}6+\cdots\,,
$$
then solve for $y$, but only partially, to get:
$$
y(1-x)=-x+\frac{x^3}6-\frac{xy^3}6+\cdots\,.
$$
Now, this tells you right away that $y$ starts out $-x+\cdots$, and since you’re only interested in things of total degree $3$ and less, you may look at the second equation and ignore the $xy^3$-term, ’cause its lowest-degree term will be $x^4/6$, ignorable. So, multiply the left-hand side of the equation by $1+x+x^2+x^3$ to get
$$
y=-x-x^2-x^3+\frac{x^3}6=-x-x^2-\frac{5x^3}6+\cdots\,.
$$
A: Notice that
$$y(x)=x\sin(y)-\sin(x)\implies y(0)=0-0=0.$$
To calculate $y'(0)$, use
$$y'(0)=\left[\frac{\mathrm{d}y}{\mathrm{d}x}\right]_{x=0}=[\sin(y)+x\cos(y)y'(x)-\cos(x)]_{x=0}=\sin(y(0))-1=-1.$$
Similarly, you can compute $y''(0)$:
$$y''(0)=[\cos(y)y'(x)+\cos(y)y'(x)-x\sin(y)(y'(x))^2+ x\cos(y)y''(x)+\sin(x)]_{x=0}=-2.$$
Finally:
$$y'''(0)=2\cos(y(0))y''(0)-0+\cos(y(0))y''(0)+\cos(0)=-4-2+1=-5.$$
In total:
$$T_3(y)=-x-\frac{2}{2!}x^2-\frac{5}{3!}x^3=-x-x^2-\frac{5}{6}x^3.$$
