Differential equation degree doubt $$\frac{dy}{dx} = \sin^{-1} (y)$$ The above equation is a form of $\frac{dy}{dx} = f(y)$, so degree should be $1$.  But if I write it as $$y = \sin\left(\frac{dy}{dx}\right)$$ then degree is not defined as it is not a polynomial in $\frac{dy}{dx}$. Please explain?
The explanation is simple: they are not the same equations. Even if two equations are equivalent, they are not exactly the same. For example:

$$\frac{dy}{dx}=\sqrt[3]{x}\tag1$$
$$\left(\frac{dy}{dx}\right)^3=x\tag2$$

The equation $(1)$ is not the same as equation $(2)$ even if they do have exactly the same solutions (in $\mathbb R$ to be clear). You can see that $(1)$ has degree $1$ and $(2)$ has degree $3$.

The problem is when you try to find degree of i.e. $$y=e^{y'}
\quad\text{or}\quad
y=\sin\left(\frac{dy}{dx}\right)\tag{a,b}$$
There exist a formula that allow you define a degree of non-polynomials, namely $$\deg\;f(x)=\lim_{x\to\infty}\frac{\log|f(x)|}{\log(x)}$$ but in some cases, such as $(b)$, is unlikely to work, whereas for other cases it allows to define a degree of non-polynomial functions. For example equation $(a)$ may be degree $\infty$.
To make sense of the definition of the order of a differential equation, you need to first isolate the highest order derivative first. For example, 
$\frac{dy}{dx} = y, \frac{dy}{dx} = y^2 + 1$ are of the first order. 
$ \frac{d^2 y}{dx^2} = y, \frac{d^2 y}{dx^2} = (\frac{dy}{dx})^3 + 1$ are of second order.

Your differential equation $\frac{dy}{dx} = \sin^{-1}y$ is first order even if it can be written in the equivalent form $y = \sin(\frac{dy}{dx}).$