dy/dx

This is an elaboration of @<> <> 's solution.

$\begin{aligned} x^y y^x & = 16 & \small \color{#3D99F6} \text{Taking natural logarithm both sides} \\ y\ln x + x \ln y & = \ln 16 & \small \color{#3D99F6} \text{Differentiating both sides} \\ \frac yx + \ln x \cdot \frac {dy}{dx} + \ln y + \frac xy \cdot \frac {dy}{dx} & = 0 & \small \color{#3D99F6} \text{Putting }x=y=2 \\ \frac 22 + \ln 2 \cdot \frac {dy}{dx}\bigg|_{x=y=2} + \ln 2 + \frac 22 \cdot \frac {dy}{dx}\bigg|_{x=y=2} & = 0 \\ 1 + \ln 2 \cdot \frac {dy}{dx}\bigg|_{x=y=2} + \ln 2 + \frac {dy}{dx}\bigg|_{x=y=2} & = 0 \\ \implies \frac {dy}{dx}\bigg|_{x=y=2} & = - \frac {1+\ln 2}{1+\ln 2} \\ & = \boxed{-1} \end{aligned}$

$\frac { dy }{ dx } =-\frac { { (F(x,y)) }_{ x }^{ ' } }{ { (F(x,y)) }_{ y }^{ ' } } =-\frac { y{ x }^{ y-1 }{ y }^{ x }\ln { y } }{ x{ y }^{ x-1 }{ x }^{ y }\ln { x } } =-\frac { { y }^{ 2 } }{ { x }^{ 2 } } \log _{ x }{ y }$

$-\frac { { y }^{ 2 } }{ { x }^{ 2 } } \log _{ x }{ y } =-\frac { 4 }{ 4 } \times 1=-1$

Take ln of both sides we get $ln{16}=yln{x}+xln{y}$ Then it's just routine implicit