a) Judging by the slope field, it would appear the solution passing through (0, 2) would just be the line [tex]y=2[/tex]. This holds up for the given ODE, since if [tex]y(x)=2[/tex], then both sides of the ODE reduce to 0.
Since we can surmise that [tex]y=2[/tex] is an equilibrium for this ODE (that is, the derivative along this line is 0 everywhere), and the slope at (1, 0) is positive, we know that as [tex]x\to+\infty[/tex], the function [tex]y(x)[/tex] will converge to 2. In other words, as [tex]x[/tex] gets larger, the slope field suggests that the solution curve through (1, 0) will start to plateau and steadily approach [tex]y=2[/tex]. On the other hand, as [tex]x\to-\infty[/tex], the slope field tells us that the curve would rapidly diverge off to [tex]-\infty[/tex]. (When you actually draw the solution, you would end up with something resembling the plot of [tex]-e^{-x}[/tex].)
b) The tangent line to [tex]y=f(x)[/tex] at [tex]x=1[/tex], given that [tex]f(1)=0[/tex], takes the form
[tex]\ell(x)=f(1)+f'(1)(x-1)[/tex]
[tex]\ell(x)=f'(1)(x-1)[/tex]
When [tex]x=1[/tex], we have [tex]y=0[/tex], so
[tex]f'(1)=\dfrac13\cdot1\cdot(0-2)^2=\dfrac43[/tex]
and so the tangent line to [tex]f(x)[/tex] at [tex]x=1[/tex] is
[tex]\ell(x)=\dfrac43(x-1)[/tex]
Using the tangent line as an approximation, we would find
[tex]f(0.7)\approx\ell(0.7)=\dfrac43\left(\dfrac7{10}-1\right)=-\dfrac4{10}=-0.4[/tex]
c) The ODE is separable, so we can write
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac13x(y-2)^2\iff\dfrac{\mathrm dy}{(y-2)^2}=\dfrac x3\,\mathrm dx[/tex]
Integrating both sides gives us
[tex]-\dfrac1{y-2}=\dfrac{x^2}6+C[/tex]
Given that [tex]y(1)=0[/tex], we get
[tex]-\dfrac1{0-2}=\dfrac{1^2}6+C\implies C=\dfrac13[/tex]
so the particular solution is
[tex]-\dfrac1{y-2}=\dfrac{x^2}6+\dfrac13[/tex]
You're asked to find the solution in the form [tex]y=f(x)[/tex], so you should solve for [tex]y[/tex]. You would end up with
[tex]y=-\dfrac6{x^2+2}+2=\dfrac{2(x^2-1)}{x^2+2}[/tex]
