A slope field is a two-dimensional image that represents the value of f(t,y) for values values of t and y for a first-order differential equation y'(t)=f(t,y(t)).
You have a third-order DE. You cannot plot a slope field.
You may be thinking of the phase portrait, which is a similar representation of the slope of an autonomous second order DE y''=f(y,y').
However, your equation is third order. The best you can do is to draw the three-dimensional phase-space for a third-order DE by converting the system into a system of first-order DEs:
$x = a(t)$
$y = a'(t)$
$z = a''(t)$
Your system becomes:
$x' = y$
$y' = z$
$z' = 2yz/x$
You can plot a unit vector in this direction in space using:
VectorPlot3D[
{y, z, 2 y z/x}/Norm[{y, z, 2 y z/x}],
{x, -3, 3}, {y, -3, 3}, {z, -3, 3},
VectorStyle -> "Segment"
]
You can adjust the viewing window accordingly, I'm guessing [-3,3] is probably not what you want. You can change VectorStyle to something else, but the arrowheads often obscure the image. Options like this:
VectorStyle -> "Arrow3D", VectorScale -> {0.2, Scaled[0.1]}
sol = ParametricNDSolve[{f'''[t] == 2 f'[t] f''[t]/f[t], f[1] == 1, f'[1] == 1, f''[1] == a}, f, {t, 1, 2}, a]; Plot[ Evaluate[Table[f'''[a][t] /. sol, {a, -2, 2, 1}]], {t, 1, 2}, PlotRange -> All]
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