human
Complete the below
def bivariate_type(f, x, y, **kwargs):

    ''''Given an expression, f, 3 tests will be done to see what type
    of composite bivariate it might be, options for u(x, y) are::
    x*y
    x+y
    x*y+x
    x*y+y
    If it matches one of these types, ``u(x, y)``, ``P(u)`` and dummy
    variable ``u`` will be returned. Solving ``P(u)`` for ``u`` and
    equating the solutions to ``u(x, y)`` and then solving for ``x`` or
    ``y`` is equivalent to solving the original expression for ``x`` or
    ``y``. If ``x`` and ``y`` represent two functions in the same
    variable, e.g. ``x = g(t)`` and ``y = h(t)``, then if ``u(x, y) - p``
    can be solved for ``t`` then these represent the solutions to
    ``P(u) = 0`` when ``p`` are the solutions of ``P(u) = 0``.
    Only positive values of ``u`` are considered.
    Examples
    >>> from sympy.solvers.solvers import solve
    >>> from sympy.solvers.bivariate import bivariate_type
    >>> from sympy.abc import x, y
    >>> eq = (x**2 - 3).subs(x, x + y)
    >>> bivariate_type(eq, x, y)
    (x + y, _u**2 - 3, _u)
    >>> uxy, pu, u = _
    >>> usol = solve(pu, u).
usol
    [sqrt(3)]
    >>> [solve(uxy - s) for s in solve(pu, u)]
    [[{x: -y + sqrt(3)}]]
    >>> all(eq.subs(s).equals(0) for sol in _ for s in sol)
    True'
    '''
assistant
#Here type
.
    u = Dummy('u', positive=True)
    if kwargs.pop('first', True):
        p = Poly(f, x, y)
        f = p.as_expr()
        _x = Dummy()
        _y = Dummy()
        rv = bivariate_type(Poly(f.subs({x: _x, y: _y}), _x, _y), _x, _y, first=False)
        if rv:
            reps = {_x: x, _y: y}
            return (rv[0].xreplace(reps), rv[1].xreplace(reps), rv[2])
        return
    p = f
    f = p.as_expr()
    args = Add.make_args(p.as_expr())
    new = []
    for a in args:
        a = _mexpand(a.subs(x, (u / y)))
        free = a.free_symbols
        if ((x in free) or (y in free)):
            break
        new.append(a)
    else:
        return ((x * y), Add(*new), u)
    def ok(f, v, c):
        new = _mexpand(f.subs(v, c))
        free = new.free_symbols
        return (None if ((x in free) or (y in free)) else new)
    new = []
    d = p.degree(x)
    if (p.degree(y) == d):
        a = root(p.coeff_monomial((x ** d)), d)
        b = root(p.coeff_monomial((y ** d)), d)
        new = ok(f, x, ((u - (b * y)) / a))
        if (new is not None):
            return (((a * x) + (b * y)), new, u)
    new = []
    d = p.degree(x)
    if (p.degree(y) == d):
        for itry in range(2):
            a = root(p.coeff_monomial(((x ** d) * (y ** d))), d)
            b = root(p.coeff_monomial((y ** d)), d)
            new = ok(f, x, (((u - (b * y)) / a) / y))
            if (new is not None):
                return ((((a * x) * y) + (b * y)), new, u)
            (x, y) = (y, x)