Simplifying $\tan^{-1} y = (\tan^{-1} x + C)$
I got this from trying to solve the differential equation $-\frac{dy}{dx} = \frac{1+y^2}{1+x^2} $
I multiplied both sides by $\tan$
$$\tan (\tan^{-1} y) = \tan(\tan^{-1} x + C) $$
for the left hand side, it is equals to $y$
But why for the right hand side, I cannot just multiply in to simplify it ? For example $ (x + \tan C) $?
why have I got to treat it like a $\tan (A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} $
Because you're not multiplying anything by "tan" -- you're undoing an operation. For example, you wouldn't "undo" performing the operation $x\mapsto\tan x$ by multiplying by $(\tan x)^{-1}=\cot x,$ which is not equal to $\tan^{-1}x$ in general. If you multiply them together you will get $1,$ obviously. But what you want to do is undo it, meaning you need to get back to what you started with, i.e. $x$. What you're actually doing is applying an 'inverse' function. In this case, $\tan^{-1}\tan x=x.$
The confusion is that the notation (something)$^{-1}$ means to 'undo' something. When you think about multiplying numbers, and that's how almost everyone learns it, the notation means to divide, since you're undoing multiplication. E.g. $2^{-1}=\tfrac12.$ But that's only one particular case of the notation. The meaning of $^{-1}$ really means to undo, and to undo something is more complicated a problem, and has a much broader meaning. So when you're talking about a function, say $f$, to undo that function is to apply another function, $f^{-1},$ which will undo whatever operation $f$ did.
So what you really want to do is this: You want to solve for $y,$ so we want to know what the 'input' $y$ is, if we know the output $\tan^{-1}y.$ So we undo that with its inverse function:
$$y=\tan(\tan^{-1}x+C).$$
Now you can see why multiplying doesn't make sense and why your next step has to be
$$y=\frac{x+\tan C}{1-x\tan C}$$