#### Answer

$$ h_{xy}(x,y) =-\frac{9x^2y^2}{(x^3+y^3)^{2}}.$$

#### Work Step by Step

Since $ h(x,y)=\ln (x^3+y^3)$, then by using the chain rule, we have
$$ h_x(x,y)=\frac{\partial h}{\partial x}=\frac{3x^2}{x^3+y^3}=3x^2(x^3+y^3)^{-1}.$$
Again using the chain rule, we get
$$ h_{xy}(x,y)=\frac{\partial^2 h}{\partial x\partial y}= -3x^2(x^3+y^3)^{-2}(3y^2)=-\frac{9x^2y^2}{(x^3+y^3)^{2}}.$$