Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
First, we need to sketch the problem, as seen below.
From the geometry of the figure below, we can see that:
$\bullet$ $\theta_1=\theta_2=\theta$
$\bullet$ $\theta_3=0^\circ$
$\bullet$ $r_1=r_2=\sqrt{x^2+d^2}$
$\bullet$ $\sin\theta=\dfrac{d}{r}=\dfrac{d}{ \sqrt{x^2+d^2}}$
$\bullet$ $\cos\theta=\dfrac{x}{r}=\dfrac{x}{\sqrt{x^2+d^2}}$
We will ignore the charge signs and focus on the electric field components' directions.
Thus, the electric field exerted by $q_1$ at a point at a distance of $x$, as seen below, is given by
$$E_1=\dfrac{kq_1}{r_1^2}\left(-\cos\theta\;\hat i+\sin\theta\;\hat j\right)$$
Plug the known;
$$E_1=\dfrac{kq}{(x^2+d^2)}\left(\dfrac{-x}{\sqrt{x^2+d^2}}\;\hat i+\dfrac{d}{ \sqrt{x^2+d^2}}\;\hat j\right)$$
$$E_1=\dfrac{kq}{\left(x^2+d^2\right)^{3/2}}\left(-x\;\hat i+d\;\hat j\right)\tag 1$$
The electric field by $q_2$ at the same point is given by
$$E_2=\dfrac{kq_2}{r_2^2}\left(-\cos\theta\;\hat i-\sin\theta\;\hat j\right)$$
Plug the known;
$$E_2=\dfrac{kq}{(x^2+d^2)}\left(\dfrac{-x}{\sqrt{x^2+d^2}}\;\hat i-\dfrac{d}{ \sqrt{x^2+d^2}}\;\hat j\right)$$
$$E_2=\dfrac{kq}{\left(x^2+d^2\right)^{3/2}}\left(-x\;\hat i-d\;\hat j\right)\tag 2$$
The electric field by $q_3$ at the same point is given by
$$E_3=\dfrac{kq_3}{r_3^2}\left(\cos0^\circ\;\hat i+\sin0^\circ\;\hat j\right)$$
Plug the known;
$$E_3=\dfrac{2kq}{x^2}\;\hat i\tag 3$$
Thus, the net electric field is given by
$$E_{net}=E_1+E_2+E_3$$
Plug from (1), (2), and (3),
$$E_{net}=\dfrac{kq}{\left(x^2+d^2\right)^{3/2}}\left(-x\;\hat i+d\;\hat j\right)+\dfrac{kq}{\left(x^2+d^2\right)^{3/2}}\left(-x\;\hat i-d\;\hat j\right)+\dfrac{2kq}{x^2}\;\hat i$$
$$E_{net}=\dfrac{-2kqx}{\left(x^2+d^2\right)^{3/2}} \;\hat i +\dfrac{2kq}{x^2}\;\hat i$$
$$E_{net}=2kq\left[ \dfrac{- x}{\left(x^2+d^2\right)^{3/2}} +\dfrac{1}{x^2}\right]\;\hat i$$
$$E_{net}=2kq\left[ \dfrac{1}{x^2}-\dfrac{ x}{\left(x^2+d^2\right)^{3/2}} \right]\;\hat i$$
And for more general cases,
$$\boxed{E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{\left(x^2+d^2\right)^{3/2}} \right]\;\hat i}$$
$$\color{blue}{\bf [b]}$$
when $x$ decreases, $x\lt \lt d$, using the boxed formula above,
$$ E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{\left(x^2+d^2\right)^{3/2}} \right]\;\hat i$$
So,
$$ E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{d^3\left((x/d)^2+1\right)^{3/2}} \right]\;\hat i$$
where $(x/d)^2\approx 0$
$$ E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{d^3 } \right]\;\hat i$$
and $x/d^3\approx 0$ as well,
$$\boxed{ E_{net}=\dfrac{2q}{4\pi \epsilon_0} \dfrac{1}{x^2} \;\hat i}$$
which is the same result if there is only charge $q_3=2q$ exerts an electric field at a distance of $x$ from it.
when $x$ increases, $x\gt \gt d$,
$$ E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{x^3\left(1+(d/x)^2\right)^{3/2}} \right]\;\hat i$$
where $(d/x)^2\approx 0$
$$ E_{net}=\dfrac{2q}{4\pi \epsilon_0}\left[ \dfrac{1}{x^2}-\dfrac{ x}{x^3 } \right]\;\hat i$$
$$\boxed{E_{net}=\bf 0\;\rm N/C}$$
which is the same result we get from a zero net charge from a point charge at a distance of $x$ fro it.
When $x$ is far away the three charges appear as a single charge of value $Q_{net}=q_1+q_2+q_3=0$.
And hence, the net electric field must be zero.
