Inspired by Satvik Golechha

Multiply through by $f(x)$ , factor, get $(f(x)-x)(f(x)-\frac1{x}) = 0$ . So for all $a$ , either $f(a) = a$ or $f(a) = 1/a$ . Note that if $f(x) = 1$ , then $x = 1$ .

If $f(a) = a$ and $f(b) = 1/b$ and $1$ is not between $a$ and $b$ , the intermediate value theorem gives a contradiction, because $1$ is between $a$ and $1/b$ , so there is a $c$ between $a$ and $b$ such that $f(c) = 1$ , so $c = 1$ , which is impossible.

This shows that if $f(a) = a$ , then $f(x) = x$ for all $x$ on the same side of $1$ as $a$ . Similarly if $f(b) = 1/b$ , then $f(x) = 1/x$ for all $x$ on the same side of $1$ as $b$ .

So there are four possibilities: $f(x) = x$ , $f(x) = 1/x$ , $f(x) = x$ for $x \le 1$ and $1/x$ for $x \ge 1$ , and $f(x) = 1/x$ for $x \le 1$ and $x$ for $x \ge 1$ . The answer is $\fbox{4}$ .

this is equivalent to, $xf^2(x)+x=x^2f(x)+f(x) \implies (f(x)-x)(xf(x)-1) =0$ This means that either , $f(x)=x$ or $f(x)=\frac{1}{x}$ . It is possible that, that for some numbers $f(x)=x$ and $f(x)=\frac{1}{x}$ . Its given that function is continuous too . So we can also have $f(x)=x, \forall x \in(0,1)$ or $f(x)=\frac{1}{x} , \forall x \in(0,1)$ and $f(x)=\frac{1}{x}, \forall x \in [1,\infty)$ or $f(x)=\frac{1}{x} , \forall x \in[1,\infty)$ respectively. This gives rise to, $2$ more functions. Hence answer is $2+2=4$