Is this a function?

x^2-2y^2-3=0

$\begin{array}{rcll} x^2-2y^2-3 &=& 0 \qquad | \qquad + 3 \\ x^2-2y^2 &=& 3 \qquad | \qquad : 3 \\ \frac{x^2}{3}-\frac{2y^2}{3} &=& 1 \\ \frac{x^2}{3}-\frac{ y^2}{\frac32} &=& 1 \\ \end{array}$

parameter notation hyperbola $\boxed{~ \begin{array}{rcll} \frac{x^2}{3}-\frac{ y^2}{1.5} &=& 1 \\ \end{array} ~}$

Yes