$\sin^2(x), \cos^2(x),$ and $\sin^4(x)$ and linear dependence Since
$$
\sin^{4}x=\sin^{2}x\sin^{2}x = \sin^{2}x(1-\cos^{2}x) = \sin^{2}x-\sin^{2}x\cos^{2}x
$$
does this mean that 
$$
\sin^{2}x,\cos^{2}x, \text{ and } \sin^{4}x
$$
are linearly dependent?

You can once again use the Wronskian: $$W(x) = \begin{vmatrix} \sin^2x & \cos^2x & \sin^4x \\ \sin(2x) & -\sin(2x) & 4\sin^3x \cos x \\ 2\cos(2x) & -2\cos(2x) & 12\sin^2x \cos^2x + -4\sin^4x\end{vmatrix},$$ and see if it is identically zero, or not. But that's quite a pain here, so you can take a leap of faith: $$W(\pi/4) = \begin{vmatrix} 1/2 & 1/2 & 1/4 \\ 1 & -1 & 1 \\ 0 & 0 & 2\end{vmatrix} = -2 \neq 0,$$ and so they are linearly independent. Or you can use the definition of linear dependence. Consider: $$a\sin^2x + b\cos^2x + c\sin^4x = 0$$
This has to be true for all $x$, so, taking $x = 0$ gives $b = 0$, leaving us with $a\sin^2x + c \sin^4x = 0$, that is, $(\sin^2x) (a + c \sin^2 x) = 0$. Now, if we don't take values of $x$ that make $\sin x = 0$, we can look only at $$a + c\sin^2x = 0.$$
Taking $x = \pi/2$ gives us $a = -c$, and $x = \pi/4$ gives $a = -c/2$. Hence $a = c = 0$ too, and the functions are linearly dependent.
