One ovious solution is x=0,y=0,z=0.
Adding the given equations pairwise, we get
2z=xyz(1a2+1b2) etc.
Hence 2xy=1a2+1b2.
Similarly,2yz=1b2+1c2,2zx=1a2+1c2.
Multiplying these, we obtain
x2y2z2=8a2b2c2(a2+b2)(b2+c2)(c2+a2)
Hence xyz=±2√2.a2b2c2√[(a2+b2)(b2+c2)(c2+a2)]
Now using the equality xy=2a2b2a2+b2, we find
z=±√2.c2√(a2+b2)√[(b2+c2)(c2+a2)]
Similarly expressions for value of x and y can be found.