Have I gone wrong with this?
$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$
Making y the subject:
$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+C}$
Can you do that with arctan?
The problem is not the arctan but that "C". Just taking the reciprocal of both sides, $\displaystyle y= \frac{1}{\frac{1}{3}arctan(\frac{x}{3})+ C}$.
Now, multiplying both numerator and denominator by 3, $\displaystyle y= \frac{3}{arctan(\frac{x}{3})+ 3C}$.
Of course, if "C" is some undetermined constant, so is 3C. You could write it as $\displaystyle y= \frac{3}{arctan(\frac{x}{3})+ C'}$
but you should understand that "C" and "C' " are not the same number.
Sorry for lateness of reply - been away at work.
That was quite sloppy of me not to see that, so:
$\displaystyle \frac{1}{y}=\frac{1}{3}arctan(\frac{x}{3})+C$
$\displaystyle y=\frac{1}{\frac{1}{3}arctan(\frac{x}{3})}+\frac{1 }{C}$
$\displaystyle y=\frac{3}{arctan(\frac{x}{3})}+\frac{3}{3C}$
$\displaystyle y=\frac{3}{arctan(\frac{x}{3})+3C}$
I gave you the explanation why it was wrong: $\displaystyle \frac{1}{a} = b + c$ does not imply $\displaystyle a = \frac{1}{b}+ \frac{1}{c}$. eg. 1/2 = 1/4 + 1/4 does not imply 2 = 4 + 4.
You are advised to review and/or take care with the basic algebra (the fact that you're working at the level of inverse trig functions means that the algebra I have explained is assumed knowledge).
$\displaystyle \frac{1}{2}=\frac{2}{4}\Rightarrow\ 1(4)=2(2)$
$\displaystyle \frac{1}{3}=\frac{2}{6}\Rightarrow\ 1(6)=2(3)$
and so on.
$\displaystyle \frac{a}{b}=\frac{c}{d}\Rightarrow\frac{b}{a}=\fra c{d}{c}$
You can use either method, but first write both sides as a single fraction.
$\displaystyle \frac{1}{y}=\frac{arctan(\frac{x}{3})}{3}+\frac{3C }{3}=\frac{arctan(\frac{x}{3})+3C}{3}$
Now you can invert the fractions (fastest) or cross-multiply
$\displaystyle 1(3)=y\left(arctan(\frac{x}{3})+3C\right)$
Now divide both sides by the multiplier of y.