Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to find the equipotentials lines. Those lines have the same potential.
We are given that the potential at a region in space is given by
$$V=100(x^2-y^2)$$
So at $V=0$,
$$x^2=y^2$$
Thus,
$$y=\pm x\tag 1$$
This means that the zero potential lines are at $\theta= \pm 45^\circ$ from $+x$-line, as you see in the first figure below.
Now we need to find the lines where $V=\pm 100\rm\; V$,
$$V=100(x^2-y^2)=100$$
So,
$$ x^2-y^2=1$$
And hence,
$$y=\pm \sqrt{x^2-1}\tag 2$$
It is obvious that this is a two-hyperbolas. One passes through $(1,0)$ and the other passes through $(-1,0)$.
Now we need to find the lines where $V=\pm 400\rm\; V$,
$$V=100(x^2-y^2)=400$$
So,
$$ x^2-y^2=4$$
And hence,
$$y=\pm \sqrt{x^2-4}\tag 3$$
It is obvious that this is a two-hyperbolas. One passes through $(2,0)$ and the other passes through $(-2,0)$.
$$\color{blue}{\bf [b]}$$
We know that the electric field is given by
$$E=-\dfrac{dV}{ds}$$
So,
$$\vec E=-\dfrac{\partial V}{\partial x}\hat i-\dfrac{\partial V}{\partial y}\hat j$$
$$\vec E=-\dfrac{\partial }{\partial x}[100(x^2-y^2)]\hat i-\dfrac{\partial V}{\partial y}[100(x^2-y^2)]\hat j$$
$$\vec E=-[-200x]\hat i-[200y]
hat j$$
$$\vec E=(\color{red}{\bf200} x \;\hat i-\color{red}{\bf 200} y\;\hat j)\;\rm V/m$$
$$\color{blue}{\bf [b]}$$
The electric field starts from positive and ends at negative lines, see the figure below.
