Find the real solutions for #{(x^4-6x^2y^2+y^4=1),(4x^3y+4xy^3=1):}#?

Subtract the second equation from the first to get:

#0 = x^4-4x^3y-6x^2y^2-4xy^3+y^4#

Divide through by #x^2y^2# to get:

#0 = (x/y)^2-4(x/y)-6-4(y/x)+(y/x)^2#

#color(white)(0)=(x/y+y/x)^2 - 4(x/y+y/x) - 8#

#color(white)(0)=(x/y+y/x)^2 - 4(x/y+y/x) + 4 - 12#

#color(white)(0)=((x/y+y/x)-2)^2-(2sqrt(3))^2#

#color(white)(0)=(x/y+y/x-2-3sqrt(2))(x/y+y/x-2+3sqrt(2))#

From the two factors, multiplying each by #(x/y)#, we get two quadratic equations in #(x/y)#...

#(x/y)^2+(-2-3sqrt(2))(x/y)+1 = 0#

#(x/y)^2+(-2+3sqrt(2))(x/y)+1 = 0#

From these we get four values for #(x/y)#:

#(x/y) = 1+(3sqrt(2))/2+-sqrt(18+12sqrt(2))/2#
(both positive)

#(x/y) = 1-(3sqrt(2))/2+-sqrt(18-12sqrt(2))/2#
(both negative)

If #(x/y) = c# then #x = cy# and:

#1 = 4x^3y+4xy^3 = 4c(1+c^2)x^4#

Hence:

#x^4 = 1/(4c(1+c^2))#

In order to have Real solutions, we require #c > 0#

So we require:

#(x/y) = 1+(3sqrt(2))/2+-sqrt(18-12sqrt(2))/2#

Due to the symmetry of the derivation in #x# and #y#, these two values are reciprocals of one another.

So we need only consider one of them then allow #x# and #y# to be swapped in the final solution...

Let #c = (x/y) = 1+(3sqrt(2))/2+sqrt(18-12sqrt(2))/2#

Then:

#x = +-root(4)(1/(4c(1+c^2)))" "# and #" "y = cx#

or:

#y = +-root(4)(1/(4c(1+c^2)))" "# and #" "x = cy#