Answer
The slope is $\ln 2 -\frac{1}{4}$.
Work Step by Step
Find $\frac{dx}{dt}$:
$\frac{dx}{dt}=\frac{d}{dt}(t^2+2t)=2t+2$
Find $\frac{dy}{dt}$:
$\frac{dy}{dt}=\frac{d}{dt}(2^t-2t)=2^t\ln 2-2$
Find $\frac{dy}{dx}$:
$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2^t\ln 2-2}{2t+2}$
Find $t$ such that $(x,y)=(15,2)$:
$t^2+2t=15$ and $2^t-2t=2$
$t^2+2t-15=0$ and $2^t-2t=2$
$(t+5)(t-3)=0$ and $2^t-2t=2$
$t=-5,t=3$ and $2^t-2t=2$
For $t=-5$, $2^{-5}-2\cdot (-5)=\frac{1}{32}+10\neq 2$ (not satisfied)
For $t=3$, $2^3-2\cdot 3=8-6=2$ (satisfied)
Find the slope at $(15,2)$:
$m=\frac{dy}{dx}=\frac{2^3\ln 2-2}{2\cdot 3+2}=\frac{8\ln 2-2}{8}=\ln 2 -\frac{1}{4}$