Answer
Tangent line: $\quad y=-2x-1$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}=-\frac{1}{3}$
Work Step by Step
The point $P$ on the curve corresponding to $t=3$ is
$\left\{\begin{array}{l}
x=-\sqrt{3+1}=-2\\
\\
y=\sqrt{3(3)}=3
\end{array}\right.\quad, P(-2,3)$
The slope of the tangent line at $t=3$ is $\left. \displaystyle \frac{dy}{dx} \right|_{t=3}$
$\displaystyle \left\{\begin{array}{l}
\frac{dx}{dt}=-\frac{1}{2}(t+1)^{-1/2},\\
\\
\frac{dy}{dt}=\frac{3}{2}(3t)^{-1/2}
\end{array}\right.\quad, \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{(\frac{3}{2})(3t)^{-1/2}}{(-\frac{1}{2})(t+1)^{-1/2}}=-\frac{3\sqrt{t+1}}{\sqrt{3t}}$
$\left. \displaystyle \frac{dy}{dx} \right|_{t=3}=\frac{-3\sqrt{3+1}}{\sqrt{3(3)}}=\frac{-3(2)}{3}=-2$
Use the point-slope equation for the tangent line.
$y-3=-2(x+2)$
$y=-2x-4+3$
$y=-2x-1$
With $y'=\displaystyle \frac{dy}{dx}$, we find $\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}$,
$\displaystyle \frac{dy'}{dt}=\frac{d}{dt}[-\frac{3\sqrt{t+1}}{\sqrt{3t}}]$
$=-\displaystyle \frac{3\cdot\frac{1}{2\sqrt{t+1}}\cdot\sqrt{3t}-(3\sqrt{t+1})\cdot\frac{1}{2\sqrt{3t}}\cdot 3}{3t}$
$=-\displaystyle \frac{3\cdot 3t-(3\sqrt{t+1})\cdot 3\sqrt{t+1}}{(3t)(2\sqrt{t+1})(\sqrt{3t})}$
$=-\displaystyle \frac{3t-(3\sqrt{t+1})\sqrt{t+1}}{2t\sqrt{t+1})(\sqrt{3t})}$
$=-\displaystyle \frac{3t-3(t+1)}{2t\sqrt{t+1}\cdot\sqrt{3t}}$
$=\displaystyle \frac{3}{2t\sqrt{t+1}\cdot\sqrt{3t}}$
$\displaystyle \frac{d^{2}y}{dx^{2}}=\frac{dy'/dt}{dx/dt}=\frac{3}{2t\sqrt{t+1}\cdot\sqrt{3t}[(-\frac{1}{2})(t+1)^{-1/2}]}=\frac{-3}{t\cdot\sqrt{3t}}$
$\displaystyle \left.\frac{d^{2}y}{dx^{2}}\right|_{t=3}=\frac{-3}{3\cdot 3}=-\frac{1}{3}$
