Solve the constraint for one variable, then substitute the result.

For our example, we solve $g=9$ for $y$, obtaining \begin{equation} y = \pm \sqrt{9-x^2} \end{equation} and insert the result into $f$, obtaining \begin{equation} f = \pm x\,\sqrt{9-x^2} \end{equation} It is now straightforward, if messy, to differentiate $f$ with respect to $x$ to find the critical points at $x=\pm\frac{3}{\sqrt2}$. Don't forget to check the endpoints at $x=\pm3$!

Parametrize the constraint, then substitute the result.

Our constraint $g=9$ is a circle, on which $x=3\,\cos\phi$, $y=3\,\sin\phi$. Substituting this parametric description of the circle into $f$ yields \begin{equation} f=9\sin\phi\cos\phi \end{equation} which can easily be differentiated with respect to $\phi$ to find the critical points at $\phi=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$. In this particular example, it is not necessary to check the “endpoints” at $\phi=0,2\pi$. (Why not?)

Use Lagrange multipliers: $\grad{f}=\lambda\grad{g}$.

In our example, we have \begin{eqnarray} \grad{f} &=& y\,\xhat + x\,\yhat \\ \grad{g} &=& 2x\,\xhat + 2y\,\yhat \end{eqnarray} so setting $\grad{f}=\lambda\grad{g}$ leads to \begin{equation} \frac{y}{2x} = \lambda = \frac{x}{2y} \end{equation} Eliminating $\lambda$ then yields $y^2=x^2$, which can be substituted into the constraint $g=9$ to determine the critical points.

The geometry behind this approach is described in § Lagrange Multipliers,

Use the cross product: $\grad{f}\times\grad{g}=\zero$.

We have already computed $\grad{f}$ and $\grad{g}$; their cross product is \begin{equation} \grad{f}\times\grad{g} = 2(y^2-x^2)\,\zhat \end{equation} which immediately yields $y^2=x^2$, as before.

Use differentials: $df=0=dg$.

In our example, we have \begin{eqnarray} df &=& y\,dx + x\,dy \\ dg &=& 2x\,dx + 2y\,dy \end{eqnarray} Solving $dg=0$ for $dy$ yields \begin{equation} dy=-\frac{x}{y}dx \label{dylag} \end{equation} and substituting this result into $df=0$ yields \begin{equation} \left( y-\frac{x^2}{y} \right) dx = 0 \end{equation} from which it again follows that $y^2=x^2$.

An equivalent description of Method 5, which brings out the geometry behind this method, is to find the points where $\grad{f}$ is perpendicular to the constraint. So we want to find the points where \begin{equation} \grad{f}\cdot d\rr = 0 \label{drlag} \end{equation} with $d\rr$ pointing along the constraint. To determine $d\rr$, we start as usual with \begin{equation} d\rr = dx\,\xhat + dy\,\yhat \end{equation} in rectangular coordinates in two dimensions, then use what we know ($g=9$) to relate $dy$ to $dx$. Thus, we solve $dg=0$ for $dy$, obtaining (\ref{dylag}), and substitute the result into $(\ref{drlag})$. But the Master Formula tells us that (\ref{drlag}) is just the statement that $df=0$ (along the curve)!

Whether one describes this method in terms of $d\rr$, which is more geometric, or simply sets $df=0$, which is more algebraic, is therefore a matter of choice; these two computations are in fact the same.

Which of these methods is used is largely a matter of personal preference. Methods 1 and 2 are the most straightforward, but also usually involve the messiest computations. Method 3 emphasizes the geometry, but the extra step of eliminating $\lambda$ seems unnecessary — unless one is interested in the value of $\lambda$ itself, as is sometimes the case in economics. Method 4 is a simplification of Method 3 for the case of one constraint, but, unlike the other methods, doesn't easily generalize to situations involving multiple constraints. Method 5 is quite robust, but requires familiarity either with $d\rr$, or at least with manipulations involving differentials.