$T(x,y,z) = (xy,yz,xz)$ For one to one, I made $(x,y,z)=(u,v,w)$ and solved.
$$xy=uv\to y=\frac{uv}{x}$$
$$\frac{uz}{x}=w$$
$$xz = uw \to x = u$$
$$uy = uv \to y = v$$
$$vz = vw \to z = w$$
So since every $(x,y,z)$ maps to a unique point $(u,v,w)$, $T(x,y,z)$ is one-to-one. Correct?
However, I'm not sure how to prove that the function is onto. I know that $T(D*) = D$, and $T\vec x = A\vec x$ where A is a $3\times3$ matrix such that $\det(A) \neq 0$. But I don't know how to work with this to solve without having some points to work with. By looking at the function, I am pretty sure that it is onto because every point in D* will map to somewhere in D. But that's not sufficient. Could I chose arbitrary points and make equations out of it and solve those? For example:
$$\begin{bmatrix}a & b & c\\d & e &f\\g &h&i\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}u\\v\\w\end{bmatrix}$$
$$ax+by+cz = u$$
$$dx+ey+fz = v$$
$$gx+hy+iz = w$$
Then choose arbitrary points for $x,y,z$ and note that from earlier $u=x,v=y,w=z$. Then solve. But, T is already given, so this may be futile. Any help?
Hint
With respect to be one-to-one:
$$T(1,0,0)=(0,0,0)=T(0,0,1)\implies \cdots$$
With respect to be onto:
Try to solve $T(x,y,z)=(-1,1,1).$ That is, $(xy,yz,xz)=(-1,1,1).$ $xz>0$ means $x$ and $z$ have the same sign. $yz>0$ means $y$ and $z$ have the same sign. Is it possible $xy<0$ if both have the same sign?