Bits of calculus, trigno and circles..what a mix

My solution's a bit... basic. It's just using elementary algebra without considering any trigonometry.

We can easily assume that $xz\ge0$ when the maximum occurs, so we'll stick to that.

I don't like subtractions, so I'll let $y=-y_1.$

Note that $x^2+{y_1}^2=9$ and $z^2+t^2=4$ to figure out that $(x^2+{y_1}^2)(z^2+t^2)=(xt+{y_1}z)^2.$

Solving this yields $xz=ty_1.$ Then we see that $ty_1\ge0$ and so,

$\sqrt{xz\cdot ty_1}\le\dfrac{xt+y_1z}{2}=3.$

Therefore, $xz\le3,$ and the maximum of $P=xz$ is $\boxed{3}.$

Similar solution to @RAHUL SINGH 's solution.

We are given:

$\begin{cases} x^2 + y^2 = 9 & \small \color{#3D99F6} \text{We can let }x = 3\cos \theta, \ y = 3\sin \theta \\ z^2 + t^2 = 4 & \small \color{#3D99F6} \text{similarly }z = 2\cos \phi, \ t = 2\sin \phi \end{cases}$

Then, we have:

$\begin{aligned} xt-yz & = 6 \\ 3\cos \theta \cdot 2\sin \phi - 3\sin \theta \cdot 2\cos \phi & = 6 \\ \sin (\phi - \theta) & = 1 \\ \phi - \theta & = \frac \pi 2 \\ \implies \phi & = \frac \pi 2 + \theta \end{aligned}$

And

$\begin{aligned} P & = xz \\ & = 3\cos \theta \cdot 2 \cos \color{#3D99F6} \phi & \small \color{#3D99F6} \text{Note that }\phi = \frac \pi 2 + \theta \\ & = 6 \cos \theta \color{#3D99F6} \cos \left(\frac \pi 2 + \theta\right) & \small \color{#3D99F6} \text{Note: }\cos x = \sin \left(\frac \pi 2-x\right) \\ & = 6 \cos \theta \color{#3D99F6} \sin \left(- \theta\right) & \small \color{#3D99F6} \text{Note: }\sin (-x) = - \sin x \\ & = {\color{#3D99F6}-} 6 \cos \theta \color{#3D99F6} \sin \theta & \small \color{#3D99F6} \text{Note: } \sin 2x = 2\sin x \cos x \\ & = - 3 \sin 2 \theta \end{aligned}$

Note that $\max (P) = \boxed{3}$ , when $\sin 2 \theta = -1$ .

since x,y,z,t lie in circles..we can assume parametric coordinates for these. let x = 3cosA, y = 3sinA, z = 2cosB, t = 2sinB... Now according to given condition xt-yz = 6....we get 6(sin(B-A)) = 6.. which leads to B = A + $\frac{Pi}{2}$ .. using this relation in the above parametric values of x,y,z,t we get xz = -3Sin(2A)...which has max. value = 3.