1. Given equation

   $2x^y+3y^x=20$

   We need to find $\dfrac{dy}{dx}$ at the point $(2,2)$.

2. Differentiate implicitly

   Recall:

   • For $x^y$, taking log helps: $x^y=e^{y\ln x}$ So, $\frac{d}{dx}(x^y)=x^y\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)$

   • For $y^x$, $y^x=e^{x\ln y}$ So, $\frac{d}{dx}(y^x)=y^x\left(\ln y+\frac{x}{y}\frac{dy}{dx}\right)$

   Differentiating $2x^y+3y^x=20$ with respect to $x$:

   $2\,x^y\left(\frac{dy}{dx}\ln x+\frac{y}{x}\right)+3\,y^x\left(\ln y+\frac{x}{y}\frac{dy}{dx}\right)=0$

3. Substitute the point $(2,2)$

   At $(x,y)=(2,2)$:

   $x^y=2^2=4, \quad y^x=2^2=4, \quad \ln x=\ln 2, \quad \ln y=\ln 2$

   Hence,

   $2\cdot 4\left(\frac{dy}{dx}\ln 2+\frac{2}{2}\right)+3\cdot 4\left(\ln 2+\frac{2}{2}\frac{dy}{dx}\right)=0$

   $8\left(\frac{dy}{dx}\ln 2+1\right)+12\left(\ln 2+\frac{dy}{dx}\right)=0$

4. Simplify

   $8\frac{dy}{dx}\ln 2+8+12\ln 2+12\frac{dy}{dx}=0$

   Grouping terms in $\dfrac{dy}{dx}$:

   $\frac{dy}{dx}(8\ln 2+12)+(8+12\ln 2)=0$

   Therefore,

   $\frac{dy}{dx}=-\frac{8+12\ln 2}{12+8\ln 2}$

   Divide numerator and denominator by $4$:

   $\frac{dy}{dx}=-\frac{2+3\ln 2}{3+2\ln 2}$

5. Convert into option form

   Using: $\ln 8=3\ln 2, \qquad \ln 4=2\ln 2$

   we get

   $\frac{dy}{dx}=-\left(\frac{2+\ln 8}{3+\ln 4}\right)$

6. Match with options

   This is exactly Option B.

Final Answer:

$\boxed{-\left(\frac{2+\log_e 8}{3+\log_e 4}\right)}$

So the correct option is B.