Introducing the so called slack variables #s_1, s_2# the optimization problem is transformed into an equivalent one
Find local minima, maxima of
#f(x,y)= x^2+y^3#
subject to
#g_1(x,y,s_1) = x + 3 x y - s_1^2 = 0#
#g_2(x.y,s_2)=x+3 x y + s_2^2-4=0#
The lagrangian is
#L(x,y,s_1,s_2,lambda_1,lambda_2) = f(x,y)+lambda_1 g_1(x,y,s_1)+lambda_2g_2(x,y,s_2)#
#L# is analytical so the stationary points include the relative maxima and minima.
The determination of stationary points is done solving for #x,y,s_1,lambda_1,s_2,lambda_2# the system of equations given by
#grad L(x,y,s_1,lambda_1,s_2,lambda_2) = vec 0#
or
#{(2 x + lambda_1 (1 + 3 y) + lambda_2 (1 + 3 y)=0),
(3 lambda_1 x + 3 lambda_2 x + 3 y^2=0),
( -s_1^2 + x + 3 x y=0),
(-2 lambda_1 s_1=0),
(-4 + s_2^2 + x + 3 x y=0)
,( 2 lambda_2 s_2=0)
:}
#
Solving we get
#(x=0., y = 0., lambda_1 = 0., s_1 = 0., lambda_2= 0., s_2= 2.)#
and
# (x= 1.12872, y= 0.847942, lambda_1=0., s_1 = 2., lambda_2 = -0.637008, s_2 = 0.)#
Both points are at the boundaries of #g_1(x,y,0)=0# and #g_2(x,y,0)=0# respectively.
Their qualification must be done with #f_(g_1)# and #f_{g_2}# respectively. So,
#f_{g_1}(x) = x^2-1/27#
#f_{g_2}(x) = -(x-4)^3/(27 x^3) + x^2#
#d^2/(dx^2)f_{g_1}(0)=2# qualifying this point as a local minimum
#d^2/(dx^2)f_{g_2}(1.12872) =11.5726# qualifying this point as a local minimum also
Attached a figure with a contour mapping with the found points
