Decouple system of SDEs / handle scaling problem Consider
$\begin{split} \newcommand{\d}{\mathrm d}
\d x &= -yx \d t + x^2 \d B\\
\d y &= -2 y^2 \d t + 2xy \d B.
\end{split}$
This is a system of two SDEs driven by the same standard Brownian motion $B$. All differentials are to be interpreted in the Ito sense.
Now we are interested in getting a dynamical description of $y$ based solely on the knowledge about $x$ and the noise $B$, i.e. we're looking for the correct functions $f, g$ s.t.
$ \mathrm d y = f(x) \mathrm d t + g(x) \mathrm d B.$
The approach I tried was via the Euler-Maruyama discretization:
$\begin{align}
x_{n+1} - x_{n} &= -h\cdot x_n \cdot y_n + \sqrt{h} \cdot x_n^2\cdot \xi_{n+1}\\
y_{n+1} - y_{n} &= -2h\cdot y_n^2 + 2\sqrt{h} \cdot x_n\cdot y_n\cdot \xi_{n+1} 
\end{align}$
Now the first equation can be solved for $y_n$:
$y_n = -\frac{x_{n+1}-x_n}{h\cdot x_n}+ \frac{x_n\cdot\xi_{n+1}}{\sqrt{h}}$
When plugging this into the r.h.s of the second line, we obtain
$y_{n+1}-y_n = -2\frac{(\Delta x_n)^2}{h\cdot x_n^2} + 2\cdot\Delta x_n \cdot\frac{\sqrt{h}\cdot \xi_{n+1}}{h}$.
Now at this point there is no problem at all: If we exchange the original iterative scheme for $y_n$ for the new one, at least numerically, we haven't done anything wrong. But we indeed have an expression for $y_{n+1}$ given only $x_n$ and the noise.
The problem arises in the limit $h\to 0$, where Deltas become differentials, $h$ becomes $\mathrm d t$ etc., so (purely informally),
$$dy = -\frac{2}{x^2}\cdot \frac{(dx)^2}{dt} + 2\frac{dx\cdot dB}{d t}.$$
Now I know that this makes no sense at all: We can't give $dx/dt$ a meaning and both terms on the r.h.s. diverge (but cancel each other out), which already happens in the equation for $y_n$.
But is there a way of making this rigorous and obtaining an expression for the dynamics of $y$ consisting only of terms in $x$ and $B$?
Note that the issue already arises in the equation
$y_n = -\frac{x_{n+1}-x_n}{h\cdot x_n}+ \frac{x_n\cdot\xi_{n+1}}{\sqrt{h}}.$
But terms on the r.h.s. are divergent for $h\to 0$ but their difference is well-behaved. 
[Sorry for crossposting this from stackexchange but my post there didn't get any response what so ever.]
 A: Alright, I found out about semimartingale decompositions and attempted the following. Can someone check for correctness?  From the defining equation of $x$ we obtain $x = -\int xy\d t + \int x^2 \d B$ which yields the unique semimartingale decomposition ($x$ is continuous) into a local martingale $M = \int x^2 \d B$ and a finite variation process $A = -\int xy\d t$. That means, $x-M$ is a process of finite variation and their derivative exists a. e. and we can write $x-M = -\int x y \d t$, hence
$$
y = -\frac{1}{x}\cdot \frac{\d(x-M)}{\d t}.
$$
Plugging this into the SDE for $y$ yields
$$
\d y = \frac{2}{x^2}\cdot \left(\frac{\mathrm d (x-M)}{\mathrm  t}\right)^2 - 2\cdot \frac{\d(x-M)}{\d t}\cdot \d B
$$
which is a valid SDE for $y$ given in terms of $x$ and $B$.
A: Here is a slightly different perspective on your decoupling question.  To recap, you asked is it possible to decouple the SDEs: $$
\begin{aligned}
&dx = - y x dt + x^2 dB \\
&dy = -2 y^2 dt + 2 x y dB 
\end{aligned}
\tag{$\star$}
$$ where $B$ is a one-dimensional standard Brownian motion.  Specifically, you wish to obtain an SDE for $y$ of the form: 
$$
dy = f(x) dt + g(x) d B
$$
In general, Ito's formula applied to $y(x)$ tells you if this decoupling is possible.  Unfortunately, in your case, it appears this decoupling is not possible.  Indeed, by Ito's formula:
$$
d y = y' dx + \frac{1}{2} x^4 y'' dt = \left( - x y y'+ \frac{1}{2} x^4 y'' \right) dt + x^2 y' dB  \tag{$\star \star$}
$$ Equating the SDE coefficients for $y$ in ($\star$) with the SDE coefficients in ($\star \star$) we obtain:
$$
x^2 y' = 2 x y \;, \quad - x y y' + \frac{1}{2} x^4 y'' = - 2 y^2
$$ These differential equations do not seem to have a solution.  To see this, suppose that $x>0$ and eliminate $y'$ from the first equation to obtain that $y''(x)=0$, which implies $y$ is a linear function of $x$. However, a linear function of $x$ cannot satisfy the first differential equation $x^2 y' = 2 x y$. 
In contrast, if you replace ($\star$) with the slightly modified SDEs
$$
\begin{aligned}
&dx = - y x dt + x^2 dB \\
&dy = - y^2 dt + 2 x y dB 
\end{aligned}
$$
and repeat the steps above, you get that $y(x) = x^2$ and
$$
d y = -x^4 dt + 2 x^3 d B
$$ However, this last SDE is a bit superfluous, since we already have an explicit formula for $y$ given $x$.
