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I'm not aware of a built-in function for a vector version of Newton's method. There might be one built into scipy that you can import. But, you can implement your own as below.

F=vector([eqx,eqy])
v0=vector([-0.34,-3.9])
print v0
for i in range(0,4):

    J=jacobian(F,[x,y]).subs(x==v0[0]).subs(y==v0[1]).n()
    v0=v0-J.inverse()*F.subs(x==v0[0]).subs(y==v0[1])
    print v0, det(J)
F.subs(x==v0[0]).subs(y==v0[1])

This gives the solution to be:

(-0.333333333333334, -3.88888888888889)