lambda and beta are non-trivial cube roots of 1 modulo n and p.
λ³ = 1 (mod n)
β³ = 1 (mod p)
[λ](x,y) = (βx,y)

Since there are 3 of each you'd have to match them (2 are non-trivial, so there is just one equality check).

You could find how to get such root of 1 here.

So, let your curve is modulo q, and has order m, both prime numbers.
Compute a primitive root of 1 modulo q:
a≠1
a^q-1 = 1 (mod q)
Then the cube root of 1 would be:
k = a^(q-1)/3 = 1^1/3 (mod q)

Instead of finding a primitive root of 1, one could directly find a cube root, this is faster:
k = c^(q-1)/3 ≠ 1 (mod q)
for c = 2,3,...

Do the same modulo the order of the curve m:
b≠1
b^m-1 = 1 (mod m)
λ = b^(m-1)/3 = 1^1/3 (mod m)

Then check which one matches:
[λ](x,y) = (kx,y)
β = k
or
[λ](x,y) = (k²x,y)
β = k² (mod q)