Pick some point $(a,b)$ which lies on the curve $y=\frac{x}{x+1}$. This means that $b=\frac{a}{a+1}$. Taking the derivative using the quotient rule, we find:
$$y'(x)=\frac{1}{(1+x)^2}$$
so the slope of the tangent line at our point $(a,b)$ is
$$y'(a)=\frac{1}{(1+a)^2}$$
The point slope formula then tells us that the equation of the tangent line to the curve at the point $(a,b)$ is
$$y-b=\frac{1}{(1+a)^2}(x-a)$$
We want the point $(1,2)$ to satisfy this equation, so remembering that $b=\frac{a}{a+1}$, we have:
$$2-b=2-\frac{a}{a+1}=\frac{1}{(1+a)^2}(1-a)$$
This equation reduces to the quadratic $a^2+4a+1=0$ (which is the quadratic you found). The solutions $a=-2\pm\sqrt{3}$ can be plugged back into the equation
$$y=\frac{1}{(1+a)^2}(x-a)+\frac{a}{1+a}=\frac{1}{(1+a)^2}x+\frac{a^2}{(1+a)^2}$$
to obtain the equations of the desired tangent lines.
Here is a helpful picture plotted with Wolfram|Alpha:
Notice where the tangent lines intersect!
Let $\epsilon>0.$ Then by Weierstrass, there is a polynomial $p$ such that $|p-h|<\epsilon/2$ on $[0,1].$
Also by Weierstrass, there is a nowhere differentiable continuous function $v$ on $[0,1].$ Let $$M=\max_{[0,1]} |v|.$$ Then $p+\epsilon v/(2M)$ is continuous and nowhere differentiable, and for $x\in [0,1],$
$$\left |\left (p(x)+\epsilon v(x)/(2M)\right)-h(x)\right|$$ $$ \le |p(x)-h(x)| + |\epsilon v(x)/(2M)|$$ $$ < \epsilon/2 + (\epsilon/2)|v(x)/M| \le \epsilon.$$
Added later: So we have shown that there is a nowhere differentiable continuous $f_\epsilon$ on $[0,1]$ such that $|f_\epsilon-h|<\epsilon$ on $[0,1].$ This implies that on $[0,1],$
$$-\epsilon < f_\epsilon < 1+\epsilon\implies 0 < f_\epsilon+\epsilon < 1+2\epsilon $$ $$\implies 0<\frac{f_\epsilon+\epsilon}{1+2\epsilon}<1.$$
Verify that the last function is continuous, nowhere differentiable, maps $[0,1]$ to $[0,1],$ and on $[0,1],$
$$\left |\frac{f_\epsilon+\epsilon}{1+2\epsilon}-h\right| < 4\epsilon.$$
Best Answer
Expanding on Peter's comments, the derivative of $f(x)=\sqrt{1-x^2}$ can indeed be found using the chain rule. For more background on derivatives if you'd like it, see here. The chain rule states that, for two functions $g(x)$ and $h(x)$, the derivative of their composition $g(h(x))$ is given by
$$\big[g(h(x))\big]'=g'(h(x))\cdot h'(x)$$
In this case, we have $g(x)=\sqrt{x}$ and $h(x)=1-x^2$. Applying the chain rule, we have
\begin{align*} g'(x)&=(x^{1/2})' \\ &=\frac{1}{2}x^{-1/2} \\ &=\frac{1}{2\sqrt{x}} \\ \implies g'(h(x))&=\frac{1}{2\sqrt{1-x^2}} \\ h'(x)&=(1-x^2)' \\ &=-2x \end{align*}
Pulling these derivatives together, we find that
\begin{align*} f'(x)&=\big[g(h(x))\big]' \\ &=g'(h(x))\cdot h'(x) \\ &=\frac{1}{2\sqrt{1-x^2}}(-2x) \\ &=\frac{-x}{\sqrt{1-x^2}} \end{align*}
We can now set the function $f(x)$ equal to its derivative $f'(x)$ to see where the two intersect:
\begin{align*} f(x)&=f'(x) \\[0.5ex] \sqrt{1-x^2}&=\frac{-x}{\sqrt{1-x^2}} \\[0.5ex] 1-x^2&=-x \end{align*}
Rearranging this, we end up with
$$x^2-x-1=0$$
This can be solved with the quadratic formula:
\begin{align*} x&=\frac{1\pm\sqrt{1-4(-1)(1)}}{2} \\ &=\frac{1\pm\sqrt{5}}{2} \end{align*}
Notice that this gives two solutions, whereas the two graphs you showed only intersect once. That's because the solution $x=\frac{1+\sqrt{5}}{2}$, when substituted back into $f(x)$, returns the square root of a negative number, which isn't a real solution. The solution $x=\frac{1-\sqrt{5}}{2}$ is therefore the only real solution. Even though the golden ratio, $\phi$, is normally given as $\phi=\frac{1+\sqrt{5}}{2}$, note that this solution we obtained is exactly $\frac{-1}{\phi}$:
$$\frac{-1}{\phi}=\frac{-2}{1+\sqrt{5}}\cdot\frac{1-\sqrt{5}}{1-\sqrt{5}}=\frac{-2(1-\sqrt{5})}{1-5}=\frac{-2(1-\sqrt{5})}{-4}=\frac{1-\sqrt{5}}{2}$$
This means that all three curves you graphed do cross at the same point! There might exist some kind of intuitive explanation as to why the intersection of a circle with its derivative involves the golden ratio, but I don't know what that explanation would be, unfortunately.