You simply replace all the $\frac{\mathrm{d}y}{\mathrm{d}x}$ terms in your second derivative with the expression you got for $\frac{\mathrm{d}y}{\mathrm{d}x}$ through implicit differentiation.
As an explicit example, suppose we wanted to find $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ for the expression $x^2+y^2-r=0$ ($r$ here is a constant). We find
$$\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{x}{y}$$
and differentiating again gives
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}{y}$$
If we substitute the expression we got for the first derivative into the second derivative, we get
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{1+\left(-\frac{x}{y}\right)^2}{y}$$
which gives
$$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}=-\frac{x^2+y^2}{y^3}$$