From inspection, I believe $\sin^3 x+\sin^3 y$ is strictly quasiconvex on the domain $x,y\in(0,\pi/2)$.
Clearly, $\sin^3x$ is strictly quasiconvex as $\sin^3(x\lambda+(1-\lambda)y)<\max(\sin^3 x, \sin^3 y)$. However, I'm having a hard time extending this to the two variable case. It appears to me that this function passes the basic test, but it fails the bordered Hessian test. Would definitely appreciate any advice on whether this is true or I am missing something. Thanks
It is not quasiconvex. One can tell from the contour plot (the contours in the upper right curve the wrong way):
For a concrete counterexample, consider the points $(\frac\pi2,\frac\pi6)$ and $(\frac\pi6,\frac\pi2)$ and their midpoint $(\frac\pi3,\frac\pi3)$: $$ \sin^3\tfrac\pi2+\sin^3\tfrac\pi6 = \sin^3\tfrac\pi6+\sin^3\tfrac\pi2 = \tfrac98 < 1.13 \quad\text{but}\quad \sin^3\tfrac\pi3+\sin^3\tfrac\pi3 = \tfrac{3\sqrt3}4 > 1.29. $$ (Of course, by continuity one can find nearby counterexamples that stay within the open square.)