Is there a way to optimise my Perlin noise?
Your check

if (array[x, y] &lt;= -0.2)
    color = Color.DarkBlue;


serves no purpose, if your number is indeed lessthan or equal -.2 then it will also hit the next check if it is lessthan or equal to 0.



A prettier way to write if-else blocks like yours would have been to use the ternary operator...

color = array[x, y] &lt;= 0.0  ? Color.DarkBlue       :
        array[x, y] &lt;= 0.1  ? Color.Blue           :
        array[x, y] &lt;= 0.2  ? Color.Beige          :
        array[x, y] &lt;= 0.22 ? Color.LightGreen     :
        array[x, y] &lt;= 0.40 ? Color.Green          :
        array[x, y] &lt;= 0.75 ? Color.DarkGreen      :
        array[x, y] &lt;= 0.8  ? Color.LightSlateGray :
        array[x, y] &lt;= 0.9  ? Color.Gray           :
        array[x, y] &lt;= 1.0  ? Color.DarkSlateGray  :
                              Color.DarkSlateGray  ;




Your method draw is a public method, and should be PascalCase not camelCase



Your worst case scenario for the if-else chain is that you have to hit every single one of those IFs. 
If you structured those in binary search design, ie. start from the middle and eliminate half of the spectrum at a time and work your way out, worst case scenario is that you have to hit half of your IFs. The code for this may look ugly, but it would help. 



The most inefficient thing about your code though, and the reason it is taking so long to process is most likely your quadruple nested for-loop

for (int y = 1; y &lt; resolutieY; y++)
{
    for (int x = 1; x &lt; resolutieX; x++)
    {
        //color calculations

        for (int i = 0; i &lt; tileSize; i++)
        {
            for (int j = 0; j &lt; tileSize; j++)
            {
                afbeelding.SetPixel(((x - 1) * tileSize) + i, ((y - 1) * tileSize) + j, color);
            }
        }
    }
}


I would recommend instead, to create the 2d array of colors in that first double for-loop, then, loop over that Color[,]  and set the pixels. 

If possible, instead just create a max resolution version of your image, and rely on the PictureBox's ability to scale the image for you.