= Solution
On the unit circle the constraint <gradient> $(2x,2y)$ never vanishes. The <Lagrange multiplier> equations are
$$
8x-3y=2\lambda x,\qquad -3x=2\lambda y,\qquad x^2+y^2=1.
$$
A nonzero solution of the first two equations requires
$$
\det\begin{pmatrix}8-2\lambda&-3\\-3&-2\lambda\end{pmatrix}
=4\lambda^2-16\lambda-9=0,
$$
so $\lambda=9/2$ or $\lambda=-1/2$. Multiplying the first equation by $x$, the second by $y$, and adding shows $2f=2\lambda(x^2+y^2)=2\lambda$. Thus each candidate value of $f$ equals its multiplier.
For $\lambda=9/2$, $y=-x/3$, giving $(x,y)=\pm(3,-1)/\sqrt{10}$. For $\lambda=-1/2$, $y=3x$, giving $(x,y)=\pm(1,3)/\sqrt{10}$. Since a continuous function attains its extrema on the <compact set> given by the <unit circle> and these exhaust the <Lagrange multiplier> candidates,
$$
\boxed{f_{\max}=\frac92\text{ at }\pm\frac{(3,-1)}{\sqrt{10}},\qquad
f_{\min}=-\frac12\text{ at }\pm\frac{(1,3)}{\sqrt{10}}.}
$$
Each $\pm$ changes both coordinates together.
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