Express this curve in the rectangular form 
Express the curve $r = \dfrac{9}{4+\sin \theta}$ in rectangular form. 
  And what is the rectangular form?

If I get the expression in rectangular form, how am I able to convert it back to polar coordinate?


what is the rectangular form?

It is the  $y=f(x)$ expression of the curve in the $x,y$ referential (see picture). It can also be the implicit form $F(x,y)=F(x,f(x))\equiv 0$. 
Steps: 
1) transformation of polar into rectangular coordinates (also known as Cartesian coordinates) (see picture)
$$x=r\cos \theta ,$$
$$y=r\sin \theta ;$$

2) from trigonometry and from 1) $r=\sqrt{x^2+y^2}$
$$\sin \theta =\frac{y}{r}=\frac{y}{\sqrt{
x^{2}+y^{2}}};$$
3) substitution in the given equation $$r=\frac{9}{4+\sin \theta }=\dfrac{9}{4+\dfrac{y}{\sqrt{x^{2}+y^{2}}}}=9\dfrac{\sqrt{x^{2}+y^{2}}}{4\sqrt{x^{2}+y^{2}}+y};$$
4) from 1)  $r=\sqrt{x^2+y^2}$, equate
$$9\frac{\sqrt{x^{2}+y^{2}}}{4\sqrt{
x^{2}+y^{2}}+y}=\sqrt{x^{2}+y^{2}};$$
5) simplify to obtain the implicit equation
$$4\sqrt{x^{2}+y^{2}}+y-9=0;$$
6) Rewrite it as $$4\sqrt{x^{2}+y^{2}}=9-y,$$ square it (which may introduce extraneous solutions, also in this question), rearrange  as
$$16y^{2}+18y+15x^{2}-81=0,$$
and solve for $y$
$$y=-\frac{3}{5}\pm \frac{4}{15}\sqrt{81-15x^{2}}.$$
7) Check for extraneous solutions. 

if I get the expression in rectangular form, how am I able to convert
  it back to polar coordinate?

The transformation of rectangular to polar coordinates is
$$r=\sqrt{x^{2}+y^{2}}, \qquad \theta =\arctan \frac{y}{x}\qquad \text{in the first quadrant},$$
or rather $\theta =\arctan2(y,x)$ to take into account a location different from the first quadrant. (Wikipedia link).
As commented by J.M. the curve is an ellipse. Here is the plot I got using the equation $16y^{2}+18y+15x^{2}-81=0$.

