The limit is 0 if you work on the real numbers, but what about the complex numbers? Try

Plot[Abs[(y^2 Sin[x])/(
   x^2 + y^2) /. {x -> t + t^3, y -> I t}], {t, -1, 1}]

I don't know if Wolfram|Alpha had this in mind, though.

Sorry, i got confused. That example was for a different purpose: it has a minimum at the origin along all straight lines, but not along parabolas. Try this:

Plot3D[(y/(x^2) )/(1 + (y/x^2)^2), {x, -1, 1}, {y, -1, 1}]

This has no limit at the origin, but

(y/(x^2) )/(1 + (y/x^2)^2) == (x^2 y)/(x^4 + y^2)

so that it has limit 0 along all radiuses. I am in a hurry, I hope I didn't make mistakes...

It is not enough that the limit is independent of the angle. You also need that the limit is uniform in the angle. A classic couterexample was found by Peano:

(x^2 - y) (x^2 - y/3)

It has limit 0 along all radiuses, but not along parables.

When u r dealing with limits of multi variable function u need to simultaneously approach the point with all variables, which means Sqrt[(x-x0)^2+(y-y0)^2]->0, when u r using Limit[Limit[,x->0],y->0] u get a repeated limit which actually has nothing to do with the existance of the limit at the discussed point, for example check the function

f(x,y)=xSin[1/x] for (x,y) not equal to (0,0) and f(0,0)=0.

At present it seems that Mathematica only makes limits in one variable. Wolfram|Alpha apparently can handle more general limits, perhaps assuming complex variables. Iterated limits or directional limits are not the same as a full two-variable limit. Wolfram|Alpha's answer is so vague that it may simply mean that it can't do it.

With complex numbers it is not bounded by 1... but I think by default Mathematica (Wolfram Language) does the limits with Real numbers... I could be wrong though...

This limit is very simple to calculate: (y^2 )/(x^2+y^2) is bounded by 1, and Sin x goes to 0. My students use Alpha to check their work and in this case Alpha got it wrong... ):