The function, its gradient and is Hessian are are implemented by the following Haskell functions:

```haskell

f::Double->Double->Double

fxy=exp(sin(50*x))

+sin(60*expy)

+sin(70*sinx)

+sin(sin(80*y))

-sin(10*(x+y))

+(x^2+y^2)/4

gradf::Double->Double->[Double]

gradfxy=[dfx,dfy]

wheredfx=50*cos(50*x)*exp(sin(50*x))

+70*cosx*cos(70*sinx)

-10*cos(10*(x+y))

+x/2

dfy=60*expy*cos(60*expy)

+80*cos(80*y)*cos(sin(80*y))

-10*cos(10*(x+y))

+y/2

hessf::Double->Double->[[Double]]

hessfxy=[[hfxx,hfxy],[hfxy,hfyy]]

wherehfxx=2500*cos(50*x)^2*exp(sin(50*x))

-2500*sin(50*x)*exp(sin(50*x))

-70*sinx*cos(70*sinx)

-4900*cosx^2*sin(70*sinx)

+100*sin(10*(x+y))

+1/2

hfyy=60*expy*cos(60*expy)

-3600*expy^2*sin(60*expy)

-6400*sin(80*y)*cos(sin(80*y))

-6400*cos(80*y)^2*sin(sin(80*y))

+100*sin(10*(x+y))

+1/2

hfxy=100*sin(10*(x+y))

```

Try to find the global minimum of the function by solving the equation

$$\nabla f(x, y) = 0$$

using e.g. Newton's method. Since the function oscillates very quickly and therefore has multiple local extrema, you should solve the equation for a sufficiently large number of initial values.

It may also be useful to try to refine the set of initial points until some heuristic criterion is fulfilled. Try to formulate such a criterion.

For reference, the global minimum is -3.306868647.