If you have a formula for $\frac{dy}{dx}$ and you want to find the value of $y'(0)$, then first you plug in $x=0$ to the implicit formula defining $y$, and evaluate to get the value of $y$. Then you plug in the value of $y$ you found and $x=0$ into the formula for $\frac{dy}{dx}$ to get the value of $y'(0)$. Similarly for $y''(0)$.
For example, suppose you are looking at the circle of radius $1$, $x^2+y^2=1$. Differentiating implicitly, we have:
\begin{align*}
\frac{d}{dx}(x^2+y^2) &= \frac{d}{dx}1\\
2x + 2y\frac{dy}{dx} &= 0\\
2y\frac{dy}{dx} &= -2x\\
\frac{dy}{dx} &= -\frac{x}{y}.
\end{align*}
To find $y'(0)$, we first plug in $x=0$ into $x^2+y^2=1$ to find the value of $y$ that corresponds to $x=0$; we get $y^2=1$, so $y=\pm 1$. Thus, we have two points, $(0,1)$ and $(0,-1)$. Then we plug in these values into the formula for $\frac{dy}{dx}$, and we get that in both cases we have $y'(0) = -\frac{0}{\pm 1}=0$. So $y'(0) = 0$.
For the second derivative, we again take derivatives:
\begin{align*}
\frac{dy}{dx} &= -\frac{x}{y}\\
\frac{d}{dx}\left(\frac{dy}{dx}\right) &= -\frac{d}{dx}\left(\frac{x}{y}\right)\\
\frac{d^2y}{dx^2} &= -\left(\frac{y(x)' - x(y')}{y^2}\right)\\
\frac{d^2y}{dx^2} &= \frac{x\frac{dy}{dx} - y}{y^2}\\
\frac{d^2y}{dx^2} &= \frac{x\left(-\frac{x}{y}\right) - y}{y^2}.
\end{align*}
The last equality, because $\frac{dy}{dx}=-\frac{x}{y}$. So now we plug in $x=0$, $y=\pm 1$ into this equation, and we get that $y''(0) = -\frac{y}{y^2} = -\frac{1}{y} = -\frac{1}{\pm 1}=\mp 1$. That is, $y''(0)=1$ on $(0,-1)$, and $y''(0)=-1$ on $(0,1)$.
Of course, if your implicit curve has a single point on the vertical line $x=0$, then you only have one value of $y$ to contend with, so the process will be simpler (as far as evaluating).
But: Notice that you don't take the derivative of the function you get after evaluating at $x=0$ to get the second derivative! You have to take the entire function and only after differentiating again do you once again plug in $x=0$ and the value of $y$ in order to find $y''(0)$.
Added: If all you want is to find the values of $y''(0)$ and $y'(0)$, you can avoid some complications in the computation of $\frac{d^2y}{dx^2}$ by starting from before you solved for $y'$. In the example above, after we find that when $x=0$ we have $y=\pm 1$ and $y'(0)=0$, we can start from the second line in the computation of $y'$:
$$2x + 2y\frac{dy}{dx} = 0,$$
and after simplifying by cancelling the $2$, we can take derivatives again: just be careful with the product rule:
\begin{align*}
x + 2yy' &= 0\\
\frac{d}{dx}(x+2yy') &= 0\\
1 + 2(y'y' + yy'') &= 0\\
2(y')^2 + yy'' &= -1.
\end{align*}
Now, plugging in $x=0$, $y'(0)=0$, and $y(0)=\pm 1$, we get $\pm 1y''(0)=-1$, hence $y''(0) = \mp 1$; i.e., $y''(0)=1$ when $y=-1$ and $y''(0)=-1$ when $y=1$. Then you don't have to worry about fractions until you start solving.