(x^3)+(y^3)= 3 x y  can be rewritten as a function of two variables (x^3)+(y^3)- 3 x y. Then the level curve of this function at the level 0 is the Folium of Decartes:

ContourPlot[(x^3)+(y^3)-3 x y,{x,-4,4},{y,-4,4},Contours->{0},ContourShading->False]

Now verify that the given parametrization satisfies this equation:

FullSimplify[((x^3)+(y^3)-3 x y)/.{x-> ((3 t)/(1+(t^3))),y-> ((3 (t^2))/(1+(t^3)))}]

Now plot the parametrization:

Show[ParametricPlot[{((3 t)/(1+(t^3))),((3 (t^2))/(1+(t^3)))},{t,-0.9,100}],ParametricPlot[{((3 t)/(1+(t^3))),((3 (t^2))/(1+(t^3)))},{t,-100,-1.1}],PlotRange->{{-4,4},{-4,4}}]

Now combine the contour plot and the parametrization plot. One can play with the domainof the parameter t to get the desired part of the Folium of Decartes.

Show[ContourPlot[(x^3)+(y^3)-3 x y,{x,-4,4},{y,-4,4},Contours->{0},ContourShading->False],ParametricPlot[{((3 t)/(1+(t^3))),((3 (t^2))/(1+(t^3)))},{t,0,1}]]

{((3 t)/(1+(t^3))),((3 (t^2))/(1+(t^3)))}

FullSimplify[D[{((3 t)/(1+(t^3))),((3 (t^2))/(1+(t^3)))},t]]

The area is equal to the following line integral:

Integrate[{0,((3 t)/(1+(t^3)))}.{((3-6 (t^3))/(((1+(t^3)))^2)),-((3 t (-2+(t^3)))/(((1+(t^3)))^2))},{t,0,Infinity}]

ContourPlot[(x^3)+(y^3)-3 x y,{x,-1.75,1.75},{y,-1.75,1.75},Contours->{0},ContourShading->False,GridLines->Automatic]