<s>[INST] Different answers of same differentiation Question with two different methods 
Question:- Find $\frac{dy}{dx}$ if $$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$

$$\arccos\bigg({\frac{x^2-y^2}{x^2+y^2}}\bigg)=\arctan( a)$$
$$\implies  \frac{x^2-y^2}{x^2+y^2}=\cos(\arctan(a))$$

Taking derivative on both sides w.r.t $\space x$, we get $\frac{dy}{dx}=\frac{y}{x}$

Out of fun, I tried the substitution $y^2=x^2\cos(\theta)$ and got
$$\frac{1-\cos(\theta)}{1+\cos(\theta)}=\cos(\arctan(a))$$ for $x\ne 0$
$$\implies \tan^2{\frac{\theta}{2}}=\cos(\arctan(a))$$
Differentiate both sides w.r.t $\space \theta$, we get
$$\tan{\frac{\theta}{2}}\bigg(1+\tan^2{\frac{\theta}{2}}\bigg)=0$$
$$\implies \tan{\frac{\theta}{2}}=0$$
$$\cos(\theta)=\frac{1-\tan^2{\frac{\theta}{2}}}{1+\tan^2{\frac{\theta}{2}}}=1$$

As $y^2=x^2\cos(\theta)$,this gives $y^2=x^2$ and $\frac{dy}{dx}=\pm 1$ Which contradicts the other method, So What's wrong with the $2^{nd}$ method?

 [/INST] In more detail, suppose that for some constant $c$
$$ c =\frac{x^2-y^2}{x^2+y^2}. \tag{1} $$
Note that this is undefined iff $\,x^2+y^2 = 0.\,$
Solve this for $\,y\,$ explicitly to get
$$ y = \pm \sqrt{\frac{1-c}{1+c}}x = \pm C\, x. \tag{2}$$
Take the implicit differential of equation $(1)$ to get
$$ 0 = \frac{4 x y (x\, dy - y\, dx)}{(x^2+y^2)^2}. \tag{3}$$
This implies that $\,\frac{dy}{dx}=\frac{y}{x}\,$
assuming that $\,x^2+y^2\ne 0\,$ and $\,x\ne 0.\,$
Note that $\,\frac{dy}{dx}=-1\,$ is true if $\,y=-x,\,$
but even in this case, $\, \frac{dy}{dx}=\frac{y}{x}\,$
so there is no contraction.
In other words, if $\,y = \pm C\,x,\,$ then
$\, \frac{dy}{dx} = \pm C = \frac{y}{x}. $
By the way, you can't

Differentiate both sides w.r.t $\,\theta$

because the explicit solution in equation $(2)$ implies that
$\,\theta\,$ is a constant.
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