There is no need for the Radon-Nikodým theorem.
By the very definition of the Radon-Nikodým derivative, we are looking for a function $g: (0,\infty) \to [0,\infty)$ which is measurable with respect to $\mathcal{A}$ and satisfies $$\lambda(E) = \int_E g \, dm, \qquad E \in \mathcal{A}. \tag{1}$$
Note that $g(x) := f(x) :=2x^2$ is not measurable with respect to $\mathcal{A}$ and therefore we cannot simply choose $g=f$.
If we define
$$E_n := \begin{cases} \bigg( \frac{1}{n+1}, \frac{1}{n} \bigg], & n \in \mathbb{N}, \\ (1,\infty), & n = 0 \end{cases}$$
then $\mathcal{A} = \sigma(E_n; n \in \mathbb{N}_0)$. Since the intervals $E_n$, $n \in \mathbb{N}_0$, are disjoint and cover $(0,\infty)$, equation $(1)$ is equivalent to
$$\lambda (E_n) = \int_{E_n} g \, dm \qquad \text{for all} \, \, n \in \mathbb{N}_0. \tag{2}$$
Moreover, any $\mathcal{A}$-measurable function $g$ is of the form
$$g(x) = \sum_{n \in \mathbb{N}_0} c_n 1_{E_n}(x) \tag{3}$$
for constants $c_n \in \mathbb{R}$. The only thing which we have to do is to choose the constants $c_n \geq 0$ such that $(2)$ holds. To this end, we plug our candidate $(3)$ into $(2)$ and find
$$\lambda(E_n) \stackrel{!}{=} \int_{E_n} g \, dm \stackrel{(3)}{=} c_n \int_{E_n} \, dm= c_n m(E_n) = c_n \left( \frac{1}{n}-\frac{1}{n+1} \right)$$
which implies
$$c_n = \lambda(E_n) n (n+1)$$
for all $n \in \mathbb{N}$. $\lambda(E_n)$ can be calculated explicitly using the very definition of $\lambda$; I leaves this to you. For $n=0$ we get $$\lambda(E_0) = \infty \stackrel{!}{=} c_0 m(E_0) = c_0 \infty,$$ i.e. we can choose $c_0 := 1$. Hence,
$$g(x) = 1_{E_0}(x)+ \sum_{n \geq 1} \lambda(E_n) n (n+1) 1_{E_n}(x)$$
is a non-negative $\mathcal{A}$-measurable function which satisfies $(2)$ (hence, $(1)$), i.e.
$$g = \frac{d\lambda}{dm}.$$
Best Answer
Yes, the Radon-Nikodym derivative is $1_{X\setminus\{0\}}$. Note that $1_{X\setminus\{0\}}\ne1$ ($\mu$-a.e.) as $0$ is an atom for $\mu$.