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.