Needs["Graphics`PlotField`"]

\(\(PlotVectorField[{1, 3  t^2/\((3  y^2 - 4)\)}, {t, 0, 
          3}, {y, \(-1.5\), 1.5}, \ Axes -> \ True, 
	
        ScaleFunction -> \((1 &)\), Ticks -> None, 
        Frame -> True, AspectRatio -> 1, \ 
        PlotPoints\  -> \ 15];\)\)

This is the direction field for the differential equation
in problem 13 on page 422.  Note that the equation can be
solved by hand to give that

expr =  y^3  - 4  y  -  t^3

is equal to a constant.  So contour plots of this equation are solutions of the original
differential equation:

ContourPlot[expr, {t, 0, 3}, {y, -1.5, 1.5}, 
      ContourShading -> False]

In particular, "expr" has value -1 at the give initial point
(1,0).  We can look at just this contour:

ContourPlot[expr, {t, 0, 3}, {y, -1.5, 1.5}, 
      Contours -> {-1}, ContourShading -> False]

This is why Euler's method freaks out somewhere around 
t=1.6.  The domain of the solution simply ends around that
point.