The original question was to find $y$ if $x \in (0,1)$, $y \in (\pi/4,\pi/2)$ and $$\arctan{(x+h)} = \arctan{x} + h\sin^2y - h\frac{\sin^2y\sin2y}{2} + h\frac{\sin^3y\sin3y}{3} - ...$$ I moved $\arctan{x}$ on to the LHS, divided by $h$ and took $\lim \limits_{h \rightarrow 0}$ on both sides and got $$\frac{1}{1+x^2} = \lim \limits_{h \rightarrow 0} \sin^2y - \frac{\sin^2y\sin2y}{2} + \frac{\sin^3y\sin3y}{3} - ...$$ Now can we find $y$ in terms of $x$ and $h$?
Will this approach work or is there another way to solve this problem?
Hint:
$$\sin^ky\sin ky=\Im\left(\sin^ky\,e^{iky}\right)$$ so that the series yields
$$\Im\left(\ln\left(1+\sin y\,e^{iy}\right)\right)=\arctan\left(\frac{\sin^2y}{1+\sin y\cos y}\right).$$
Then
$$\frac{\sin^2y}{1+\sin y\cos y}=\tan\left(\frac{\arctan(x+h)-\arctan x}{h}\right)=:t,$$ which can be solved for $y$.
You can rewrite as
$$2-\cos(2y)=t(2+\sin(2y))$$ to get a linear trigonometric equation that has a well-know solution. The expression of $t$ des not simplify.