$f(x)= \frac{quadratic expression}{quadratic expression}$
$f(x)= \frac{linear function}{quadratic expression}$
$f(x)= \frac{quadratic expression}{linear function}$
Above 3 functions are always many to one, given numerator and denominator do not have any common factor.
Is there any (not necessarily formal) proof of this (undergraduate level)
Let $f(x) = k$ for some $x$.
Consider case (1) as it subsumes the others. If $f(x) = \frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2}$ Then \begin{align*} \frac{a_1x^2+b_1x+c_1}{a_2x^2+b_2x+c_2} &= k\\ a_1x^2+b_1x+c_1&= ka_2x^2+kb_2x+kc_2\\ (a_1-ka_2)x^2+(b_1-kb_2)x+(c_1-kc_2)&= 0\\ \end{align*} This is a quadratic in x that has 0 solutions (in $\mathbb R$), 1 or 2 solutions, depending on whether $(b_1-kb_2)^2 - 4 (a_1-ka_2)(c_1-kc_2)$ is less than 0, equal to 0, or more than 0 respectively. Call that expression $g(k)=\text{discriminant}(f(x))$ \begin{align*} g(k)&=(b_1-kb_2)^2 - 4 (a_1-ka_2)(c_1-kc_2)\\ &= (k^2b_2^2+b_1^2-2kb_1b_2)-4(k^2a_2c_2+a_1c_1-k(a_2c_1+a_1c_2))\\ &= k^2(b_2^2-4a_2c_2)+2k(2a_2c_1+2a_1c_2-b_1b_2)+(b_1^2-4a_1c_1)\tag{0}\\ \end{align*}
If $x$ has no solutions in $\mathbb R$, then $g(k)$ has to lie entirely below the reals for all k.
Again we get the condition that the coefficient of $k^2$ in $g(k)$ is negative, and it has no solutions.
Therefore we have \begin{align*} b_2^2 - 4a_2c_2&<0\tag 1\\ \end{align*} and \begin{align*} (2(2a_2c_1+2a_1c_2-b_1b_2))^2-4(b_2^2-4a_2c_2)(b_1^2-4a_1c_1)&\stackrel{?}{<}0\\ (2a_2c_1+2a_1c_2-b_1b_2)^2-(b_2^2-4a_2c_2)(b_1^2-4a_1c_1)&\stackrel{?}{<}0\\ (4a_2^2c_1^2+4a_1^2c_2^2+b_1^2b_2^2+8a_1a_2c_1c_2-4a_2c_1b_1b_2-4a_1c_2b_1b_2)-\\(b_1^2b_2^2 +16a_1a_2c_1c_2-4a_2c_2b_1^2-4a_1c_1b_2^2)&\stackrel{?}{<}0\\ 4a_2^2c_1^2+4a_1^2c_2^2-8a_1a_2c_1c_2-4a_2c_1b_1b_2-4a_1c_2b_1b_2+4a_2c_2b_1^2+4a_1c_1b_2^2&\stackrel{?}{<}0\\ 4(a_1c_2-a_2c_1)^2-4a_2c_1b_1b_2-4a_1c_2b_1b_2+4a_2c_2b_1^2+4a_1c_1b_2^2&\stackrel{?}{<}0\\ 4(a_1c_2-a_2c_1)^2-4(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)&\stackrel{?}{<}0\\ (a_1c_2-a_2c_1)^2-(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)&\stackrel{?}{<}0\\ \end{align*} Consider vectors $a = [a_1,a_2],b = [b_1,b_2],c = [c_1,c_2]$. Then the above can be thought of as the cross products. \begin{align*} (a_1c_2-a_2c_1)^2-(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)&\stackrel{?}{<}0 \equiv (a\times c)^2 \stackrel{?}{<}(a\times b) (b\times c)\\ 0 &\stackrel{?}{<}(a\times b) (b\times c) - (a\times c)^2\tag{2} \end{align*}
Now we can consider the RHS of (2) is our optimization function and (1) is the constraint
Assume without loss of generality that $c$ lies at an angle of $\theta_{a,c}$ to $a$ where $\theta_{a,c} \in [0,\pi]$ Now note that $b$ can be in two angles, either it can be between $a$ and $c$, or it can lie between $-a$ and $-c$. Irrespective of which side it lies on, the area of the parallelograms formed by $(a,b)$ and $(b,c)$ stay the same. Therefore we can assume $b$ lies between $a$ and $c$.
Now we make two observations (i) the value of (2) is maximized under constraint (1) when $a$ and $c$ are symmetric with respect to $b$. (ii) Fixing $|a|$ and $|c|$, to maximize $a_2$ and $c_2$ for constraint (1), they are symmetric around the $y$ axis, ie $a_2=c_2$.
Specifically, (i) and (ii) taken together imply to maximize (1) with constraint (2), $b$ points "up", ie $b=(0,|b|)$, $|a| = |c|$ and $\theta_{a,b} = \theta_{b,c} = \theta$
\begin{align*} b_2^2 &< 4a_2c_2\\ |b|^2 &< 4|a|\cos(\theta)|c|\cos(\theta)\\ |b|^2 &< 4|a|\cos(\theta)|a|\cos(\theta)\\ |b|^2 &< 4|a|^2\cos^2(\theta)\\ \end{align*} Now we will use this in the optimization equation (2). \begin{align*} (a\times b) (b\times c) - (a\times c)^2&=|a||b|\sin[\theta_{a,b}]|b||c|\sin[\theta_{b,c}] - (a||c|\sin[\theta_{a,c}])^2\\ &=|a||c||b|^2\sin[\theta]\sin[\theta] - (|a||c|sin[2\theta])^2\\ &=|a||c||b|^2\sin^2[\theta] - |a|^2|c|^2sin^2[2\theta]\\ &\leq|a|^2|b|^2\sin^2[\theta] - |a|^4sin^2[2\theta]\\ &<|a|^2(4|a|^2\cos^2[\theta])\sin^2[\theta] - |a|^4sin^2[2\theta]\\ &<4|a|^4\cos^2[\theta]\sin^2[\theta] - |a|^4sin^2[2\theta]\\ &<|a|^4 (4\cos^2[\theta]\sin^2[\theta] - sin^2[2\theta])\\ &<|a|^4 (sin^2[2\theta] - sin^2[2\theta])\\ &<0\\ \end{align*}
Now we see that if $b_2^2 - 4a_2c_2 >0$, then $g(k)$ is a quadratic pointing upwards. If not, then $\text{discriminant}(g(k))>0$, which means g(k) has two roots. This shows us that there are some $k$ for which $g(k)$ is positive.
But $g(k)$ is the discriminant of $f(x)$. Therefore there are some k, for which $\text{discriminant}(f(x))$ is positive, and therefore f(x) has two real roots.
Therefore $f(x)$ is a many to one function.