Conflicting answers, chain rule Let:
$g:\mathbb{R}^2\rightarrow\mathbb{R}^2$ and $g(x,y) = (x^2-y^2,2xy)$
$f:\mathbb{R}^2\rightarrow\mathbb{R}$ and $f(u,v) = u^2+v^2$
first
compute the 2x2 matrix $(Dg)(x,y)$ (don't know why it's written like that, compute the total derivative of g at (x,y)
Easy, $\frac{\partial{g}}{\partial{x}}=(2x,2y)$ and $\frac{\partial{g}}{\partial{y}}=(-2y,2x)$
$$\begin{pmatrix}2x&&-2y\\2y&&2x\end{pmatrix}$$
second
Calculate $\nabla f(u,v) = (2u,2v)$
third
Use thee chain rule to calculate $D(f$ composed $g)(x,y)$
$D_a(f(g))=D_{g(a)}f$ composed $D_ag$ (total derivative at a point a=(x,y))
$D_{g(a)}f = \nabla f(x^2-y^2,2xy) = (2x^2-2y^2,4xy)$
Which is:
$$\begin{pmatrix}2x&&-2y\\2y&&2x\end{pmatrix}*\begin{pmatrix}2x^2-2y^2\\4xy\end{pmatrix}$$
That gives the vector:
$(2x(2x^2-2y^2-4y^2),2y(2x^2+2y^2))$ which I believe contains a sign error at the -4.
$f(g(x,y))=f(x^2-y^2,2xy)=(x^2-y^2)^2+4x^2y^2=x^4-2x^2y^2+y^4+4x^2y^2=x^2+2x^2y^2+y^4=(x^2+y^2)^2$
So $\nabla(f(g(x,y)) = (2(x^2+y^2)2x,2(x^2+y^2)2y)$
(I haven't said any more in my workings, I have just checked over and over again, you an see that, without that potential sign error, this'd work)
My notation is particularly weak here, so any help (defining what things mean) would be gratefully received for example $\nabla(f(g(x,y))$ Invites one to use the chain rule, although it denotes that I substituted. 
 A: The chain rule yields
$$\nabla(f\circ g)=\begin{pmatrix} u_x & v_x \\ u_y & v_y\end{pmatrix}\begin{pmatrix}f_u(g) \\ f_v(g)\end{pmatrix}=\begin{pmatrix} 2x & 2y \\ -2y & 2x\end{pmatrix}\begin{pmatrix}2(x^2-y^2) \\ 4xy\end{pmatrix}.$$
Albeit the usual statement of the chain rule would be $D_{g(a)}f\cdot Dg$ in which case it would be
$$\begin{pmatrix}f_u(g) & f_v(g)\end{pmatrix}\begin{pmatrix}u_x & u_y \\ v_x & v_y\end{pmatrix};$$
if this is what you were trying to do then you had the $\nabla f$ on the wrong side of the matrix.
A: I have added the component differentials on which the derivatives are to operate to get the total differential:
$$
d(f\circ g)
=\nabla \left(f\circ g\right)\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}
=\nabla f(g)\underbrace{\frac{\partial g}{\partial(x,y)}\begin{bmatrix}\mathrm{d}x\\\mathrm{d}y\end{bmatrix}}_{\mathrm{d}g}
$$
Thus, the gradient is expected to be a row vector to multiply by the column differential. This is because the gradient of a function is a covariant vector and the coordinate vector (and therefore the differential vector) is a contravariant vector.
As you computed,
$$
\frac{\partial g}{\partial(x,y)}=\begin{bmatrix}2x&&-2y\\2y&&2x\end{bmatrix}
$$
and
$$
\nabla f(g)=\begin{bmatrix}2x^2-2y^2&4xy\end{bmatrix}
$$
therefore
$$
\begin{align}
\nabla \left(f\circ g\right)
&=\begin{bmatrix}2x^2-2y^2&4xy\end{bmatrix}
\begin{bmatrix}2x&&-2y\\2y&&2x\end{bmatrix}\\
&=\begin{bmatrix}4x^3+4xy^2&4x^2y+4y^3\end{bmatrix}\\
&=4(x^2+y^2)\begin{bmatrix}x&y\end{bmatrix}
\end{align}
$$
It appears that the only error in your computations was taking the gradient as a column vector and multiplying on the wrong side.
A: The function $f(g(x,y)$ is just $(x^{2}+y^{2})^{2}$. So the total derivative is $$D_{(x,y)}=2(x^{2}+y^{2})*2x*dx+2(x^{2}+y^{2})*2y*dy$$
If you want to do it via formal chain rule, then you have to write this as 
$$D_{x}f(u,v)=\frac{\partial f}{\partial u}*\frac{\partial{u}}{\partial x}dx+\frac{\partial f}{\partial v}*\frac{\partial{v}}{\partial x}dx,D_{y}f(u,v)=\frac{\partial f}{\partial u}*\frac{\partial{u}}{\partial y}dy+\frac{\partial f}{\partial v}*\frac{\partial{v}}{\partial y}dy$$
and proceed formally. The result should be the same. 
Update:
In general a derivative means a map from the tangent space of the domain to the tangent space of the codomain. So if we have a composite map $h=f\circ g$, then the derivative of $h$ become 
$$Dh(v)=Df(Dg(v))=(Df\times Dg)(v)$$
In this particular case as you computed we have $Df=(2x^{2}-2y^{2},4xy)$, $$Dg=\begin{pmatrix}2x&&-2y\\2y&&2x\end{pmatrix}$$So the product is 
$$(2x^{2}-2y^{2},4xy)*\begin{pmatrix}2x&&-2y\\2y&&2x\end{pmatrix}=(4x^{3}+4y^{2}x,4y^{3}+4x^{2}y)$$
The confusion might be that for a differential map from $\mathbb{R}^{n}$ to $\mathbb{R}^{m}$, the Jacobian is an $m\times n$ matrix. The map $f:\mathbb{R}^{2}\rightarrow \mathbb{R}^{1}$ thus has a derivative $D_{f}$ which must be a $1\times 2$ matrix. The following links might be helpful:
http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
http://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions
