Let $OABC$ be a unit square in the $xy$-plane with $O(0,0),A(1,0),B(1,1)$ and $C(0,1)$. Let $u=x^2-y^2$, and $v=xy$ be a transformation of the $xy$-plane into the $uv$-plane. The transform (or image) of the square is:
Each point on the square can be in the form $(0,y)$, $(1,y)$, $(x,0)$, and $(x,1)$, where $0 \le x,y \le 1$.  Making the appropriate substitutions results in points being $(-y^2,0)$, $(1-y^2,2y)$, $(x^2,0)$, and $(x^2 - 1,2x)$ on the $uv$-plane.  
Notice that since $v \ge 0$, none of the points are below the u-axis, so options A,B, and E are out.  Since $x = \tfrac{v}{2}$, $u = (\tfrac{v}{2})^2 - 1$, so $v = 2\sqrt{u+1}$, where $-1 \le u \le 0$.  That means some of the lines are not straight, so the answer is $\boxed{\textbf{(D)}}$.