One more of those

Let $x = \cos(\theta)\sin(\phi), y = \sin(\theta)$ and $z = \cos(\theta)\cos(\phi).$ Then $x^{2} + y^{2} + z^{2} = 1$ and

$1330xy + 3192yz = \sin(\theta)\cos(\theta)(1330\sin(\phi) + 3192\cos(\phi)) =$

$\dfrac{\sin(2\theta)}{2}*\sqrt{1330^{2} + 3192^{2}}*\left(\dfrac{1330}{\sqrt{1330^{2} + 3192^{2}}}\sin(\phi) + \dfrac{3192}{\sqrt{1330^{2} + 3192^{2}}}\cos(\phi)\right) =$

$\dfrac{\sin(2\theta)}{2}*3458*\sin(\alpha + \phi) = 1729*\sin(2\theta)\sin(\alpha + \phi),$ where $\alpha = \cos^{-1}\left(\dfrac{1330}{3458}\right).$

This expression has a maximum value of $\boxed{1729}$ when $\theta = 45^{\circ}$ and $\phi = 90^{\circ} - \alpha.$

As noted by Harsh, the maximum in general for $axy + byz$ given that $x^{2} + y^{2} + z^{2} = 1$ is

$\dfrac{1}{2}\sqrt{a^{2} + b^{2}}.$

Probably the Harsh Shrivastava and Brian Charlesworth posted awesome solution but I will stick to the basic AM-GM.

We have

$\Large{\left( { x }^{ 2 }+\frac { { 1330 }^{ 2 }{ y }^{ 2 } }{ { 1330 }^{ 2 }+{ 3192 }^{ 2 } } \right) +\left( \frac { { 3192 }^{ 2 }{ y }^{ 2 } }{ { 1330 }^{ 2 }+{ 3192 }^{ 2 } } +{ z }^{ 2 } \right) \\ \ge 2\sqrt { { x }^{ 2 }\times \frac { { 1330 }^{ 2 }{ y }^{ 2 } }{ { 1330 }^{ 2 }+{ 3192 }^{ 2 } } } +2\sqrt { \frac { { 3192 }^{ 2 }{ y }^{ 2 } }{ { 1330 }^{ 2 }+{ 3192 }^{ 2 } } \times { z }^{ 2 } } \\ \\ 1\ge \frac { 2 }{ \sqrt { { 1330 }^{ 2 }+{ 3192 }^{ 2 } } } \left( 1330xy+3192yz \right) \\ \\ 1729\ge \left( 1330xy+3192yz \right) }$

Now for the general expression we have

$\Large{\left( { x }^{ 2 }+\frac { { a }^{ 2 }{ y }^{ 2 } }{ { { a }^{ 2 }+{ b }^{ 2 } } } \right) +\left( \frac { { b }^{ 2 }{ y }^{ 2 } }{ { { a }^{ 2 }+{ b }^{ 2 } } } +{ z }^{ 2 } \right) \\ \ge 2\sqrt { { x }^{ 2 }\times \frac { { a }^{ 2 }{ y }^{ 2 } }{ { { a }^{ 2 }+{ b }^{ 2 } } } } +2\sqrt { \frac { { b }^{ 2 }{ y }^{ 2 } }{ { { a }^{ 2 }+{ b }^{ 2 } } } \times { z }^{ 2 } } \\ \\ \frac { \sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ 2 } \ge axy+byz}$

Answer to BONUS QUESTION : $\dfrac{1}{2} \sqrt{ a^{2} + b^{2}}$

First of all let $K$ = Maximum value of $axy+byz$ subject to constraints $x^2+y^2+z^2=1$

$K = y(ax+bz)$

By Cauchy Schwarz inequality, $ax+bz \leq \sqrt{(a^{2} + b^{2})(x^{2}+ z^{2})} = \sqrt{(a^{2}+b^{2})(1 - y^{2})}$

$\implies K = y(\sqrt{(a^{2}+b^{2})(1 - y^{2})})$

$\implies K =(\sqrt{(a^{2}+b^{2})(-(y^{4} - y^{2} + (0.5)^{2}) + (0.5)^{2}})$

$\implies K = \sqrt{(a^{2}+b^{2})((\frac{1}{2}) ^{2}-(y^{2} - \frac{1}{2})^{2} })$

For maximum, y should be equal to $\sqrt{\frac{1}{2}}$ .

$\implies K =\sqrt{(a^{2}+b^{2})(\frac{1}{2})^{2} - (0)^{2} })$

$\implies K = \dfrac{1}{2} \sqrt{ a^{2} + b^{2}}$

For maximal value of axy+byz we can simplify it to y*(ax+bz).

Now, if sum of squares of x,y,z is 1 then (x^2 + z^2 = 1- y^2).....................(i)

Considering only (ax+bz), we can apply Cauchy-Schwarz Inequality.

Which is (ax+bz)^2 <= (a^2 + b^2)*(x^2 + z^2)

after taking the square root on both sides we obtain,

(ax+bz) <= sqrt [ (a^2 + b^2)*(x^2 + z^2) ]

Then by multiplying y on both sides we have.

y (ax+bz) <= y * {sqrt [ (a^2 + b^2) (x^2 + z^2) ]}

after taking y on RHS inside the square root and using (i) we obtain,

y (ax+bz) <= sqrt[ (a^2 + b^2) (y^2 - y^4)]

Now by applying calculus we can find the maximum value of the equation (y^2 - y^4) , which turns out to be 1/4.

Hence, we have max(axy+byz) <= (1/2) * sqrt(a^2 + b^2).

So for this question we have a=1330 , b=3192,

hence after evaluating we have the answer to be 1729.

This is my approach, please guide me if I am wrong somewhere. Also please know that I am not a good LATEX user.