First we find the direction vector of the line: $\mathbf{v}=\langle-1,5,6\rangle-\langle-3,1,0\rangle=\langle 2,4,6\rangle$. Therefore the line through the given points can be written as
$$
\mathbf{r}(t)=\langle-3,1,0\rangle+t\langle 2,4,6\rangle,
$$
or yet $x(t)=-3+2 t, y(t)=1+4 t, z(t)=6 t$. Plugging these equations into the equation of the plane,
$$
2(-3+2 t)+(1+4 t)-6 t+2=0
$$
gives $t=3 / 2$, thus $(0,7,9)$

Solution 4. For (a), write $y=y(x)$ so that $F(x, y)=F(x, y(x))$ and apply the chain rule:
$$
\frac{d}{d x} F(x, y(x))=\frac{\partial F}{\partial x}(x, y(x)) \frac{d x}{d x}+\frac{\partial F}{\partial y}(x, y(x)) \frac{d y}{d x}=\frac{d}{d x} 5=0,
$$
which gives the result since $\frac{d x}{d x}=1$.
(b) Implicit function theorem: If $F=F(x, y)$ is defined in a neighborhood of $\left(x_0, y_0\right), F\left(x_0, y_0\right)=$ $0, F_x$ and $F_y$ exist and are continuous, and $F_y\left(x_0, y_0\right) \neq 0$, then $F(x, y)=0$ defines $y$ as a function of $x$ in a neighborhood of $\left(x_0, y_0\right)$.

The basic intuition is the following. If $F_y\left(x_0, y_0\right) \neq 0$, then the tangent line to the curve $F(x, y)=$ 0 at $\left(x_0, y_0\right)$ is not parallel to the $y$-axis. Since this tangent line approximates the curve near $\left(x_0, y_0\right)$, there does not exist a vertical line that intersects that curve $F(x, y)=0$ twice in a small neighborhood of $\left(x_0, y_0\right)$. Hence, for each point $(x, y)$ on the curve (and near $\left.\left(x_0, y_0\right)\right)$ there corresponds a unique $y$, which means that $y$ can be viewed as a function of $x$ along $F(x, y)=0$ and near $\left(x_0, y_0\right)$.

(a) $f$ is a continuous function and $D$ is a closed bounded set. Hence $f$ attains an absolute maximum and absolute minimum in $D$ by the extreme value theorem.
(b) Compute $f_x(x, y)=4 y^2-2 x y^2-y^3=0$. This gives $y=0$ or $y=4-2 x$. Since $y=0$ belong to the boundary of $D$, which will be analyzed separately below, we can ignore it for now. Next, find $f_y(x, y)=8 x y-2 x^2 y-3 x y^2$. Setting this expression equal to zero and plugging in $y=4-2 x$ we find
$$
8 x(4-2 x)-2 x^2(4-2 x)-3 x(4-2 x)^2=x(4-2 x)(4 x-4)=0
$$
which gives $x=0, x=1$, or $x=2$. As before, $x=0$ is on the boundary of $D$ so we can ignore it for now. When $x=2$, we have $y=4-2 x=0$ which can also be ignored at this point since $y$ then belongs to the boundary of $D$. Hence, we are left with $x=1$, so $y=2$, and this gives $f(1,2)=4$.
Now we analyze the behavior of $f$ on the boundary of $D$. The boundary is given by three lines:
$$
L_1:{x=0,0 \leq x \leq 6}, L_2:{(x, y) \mid y=-x+6,0 \leq x \leq 6}, L_3:{y=0,0 \leq y \leq 6}
$$
Along $L_1$ and $L_3$ we have $f(x, 0)=0$ and $f(0, y)=0$, respectively. Along $L_2$,
$$
g(x)=f(x, 6-x)=-2\left(x^3-12 x^2+36 x\right)
$$
We seek that absolute maximum and minimum of $g$ on $[0,6]$. Computing $g^{\prime}(x)=-6(x-2)(x-6)=$ 0 we find $x=2$ and $x=6$. For $x=2$ we have $y=4$ and $f(2,4)=-64 . x=6$ is an endpoint, and the endpoints $x=0$ and $x=6$ have already been tested in $L_3$ and $L_1$, respectively.

We conclude that the absolute maximum is 4 , occurring at $(1,2)$, and the absolute minimum is $-64$, occurring at $(2,4)$