Can the slope here can be found?

Let the cubic function be $f(x)$ and its three real roots be $a$ , $b$ , and $c$ such that $a<b<c$ . Then we have:

$\begin{aligned} f(x) & = (x-a)(x-b)(x-c) \\ \implies f'(x) & = (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) \\ f'(a) & = (a-b)(a-c) = a^2 - ab + bc - ca = 2 & \small \blue{\text{Slope of the blue line}} \\ f'(c) & = (c-a)(c-b) = c^2 + ab - bc -ca = 1 & \small \pink{\text{Slope of the pink line}} \\ f'(a)+f'(c) & = a^2 + c^2 - 2ca = (c-a)^2 = 3 \\ \implies f'(b) & = (b-a)(b-c) & \small \orange{\text{Slope of the yellow line}} \\ & = - \frac {\blue{(a-b)(a-c)}\pink{(c-a)(c-b)}}\red{(c-a)^2} & \small \red{\text{Note that }(c-a)^2 = 3} \\ & = - \frac {\blue{f'(a)}\pink{f'(c)}}\red 3 \\ & = - \frac {\blue 2 \cdot \pink 1}3 \\ & = \boxed {-\frac 23} \end{aligned}$

$f(x)=A(x-a)(x-b)(x-c)$

$f\prime(x)=A(x-b)(x-c)+A(x-a)(x-c)+A(x-a)(x-b)$

$f\prime(a)=A(a-b)(a-c)$ ,

$f\prime(b)=A(b-a)(b-c)$

$f\prime(c)=A(c-a)(c-b)$

$f\prime(a)+f\prime(c)=A(a-b)(a-c)+A(c-a)(c-b)=A(a-c)(a-b-c+b)=A(a-c)^2$

$f\prime(a)f\prime(c)=A(a-b)(a-c)A(c-a)(c-b)=-A^2(a-c)^2(a-b)(c-b)=-(A(b-a)(b-c))(A(a-c)^2)=-f\prime(b)(f\prime(a)+f\prime(c))$

So $f\prime(b)=-\frac{f\prime(a)f\prime(c)}{f\prime(a)+f\prime(c)} = -\frac{1*2}{1+2}=-\frac{2}{3}$

Here are two solutions sent to me in Twitter and Instagram