Update: I the point on the surface of this function I wish to obtain is {0.2,200}.It appears from the shape of the function it is easiest to first find the y point which always seems to have the 'features' shown in the plot.

Update 1: The point on the surface of this function I wish to obtain is {0.2,200}.It appears from the shape of the function it is easiest to first find the y point which always seems to have the 'features' shown in the plot.

Update 2: The following constraints always hold: a>0 and b>0

Update: I the point on the surface of this function I wish to obtain is {0.2,200}.It appears from the shape of the function it is easiest to first find the y point which always seems to have the 'features' shown in the plot.

I have a function that is shown in the image, below. The point I am interested in extracting is, in this case, 200. This appears to be the point at which one derivative is infinite.
It is also possible this is the inflection point?

I have a function that is shown in the image, below. The point I am interested in extracting (numeric fine) is, in this case, 200. This appears to be the point at which one derivative is infinite.
It is also possible this is the inflection point?

Plot3D[f[a,b], {a, 0, 400}, {b, 0, 0.5}]

    f[a_, b_] = 
 1/b 30.` (-7.681373803360649` + 
    0.011809163818525368` a (-((
        20.60129077457011` E^(-((
          0.03333333333333333` (-4.605170185988092` + 
             15.` (0.05` - b^2/2) + Log[a])^2)/b^2)))/(a b)) + (
       0.2060129077457011` E^(
        7.5` b^2 + 15.` (0.05` - b^2/2) - (
         0.03333333333333333` (-4.605170185988092` + 15.` b^2 + 
            15.` (0.05` - b^2/2) + Log[a])^2)/b^2))/b + 
       E^(7.5` b^2 + 
         15.` (0.05` - b^2/2)) (1 + 
          Erf[(0.18257418583505533` (-4.605170185988092` + 15.` b^2 + 
              15.` (0.05` - b^2/2) + Log[a]))/b])) - 
    0.0033333333333333335` (-3.54274914555761` (-100.` (1 + 
             Erf[(0.18257418583505533` (-4.605170185988092` + 
                 15.` (0.05` - b^2/2) + Log[a]))/b]) + 
          a E^(7.5` b^2 + 
            15.` (0.05` - b^2/2)) (1 + 
             Erf[(0.18257418583505533` (-4.605170185988092` + 
                 15.` b^2 + 15.` (0.05` - b^2/2) + Log[a]))/b])) + 
       0.23618327637050734` (-((
           309.01936161855167` E^(-((
             0.03333333333333333` (-4.605170185988092` + 
                15.` (0.05` - b^2/2) + Log[a])^2)/b^2)))/b) + (
          
          3.090193616185517` a E^(
           7.5` b^2 + 15.` (0.05` - b^2/2) - (
            0.03333333333333333` (-4.605170185988092` + 15.` b^2 + 
               15.` (0.05` - b^2/2) + Log[a])^2)/b^2))/b + 
          15.` a E^(
           7.5` b^2 + 
            15.` (0.05` - b^2/2)) (1 + 
             Erf[(0.18257418583505533` (-4.605170185988092` + 
                 15.` b^2 + 15.` (0.05` - b^2/2) + Log[a]))/b]))) + 
    0.11809163818525367` a^2 b^2 ((
       20.60129077457011` E^(-((
         0.03333333333333333` (-4.605170185988092` + 
            15.` (0.05` - b^2/2) + Log[a])^2)/b^2)))/(a^2 b) + (
       0.2060129077457011` E^(
        7.5` b^2 + 15.` (0.05` - b^2/2) - (
         0.03333333333333333` (-4.605170185988092` + 15.` b^2 + 
            15.` (0.05` - b^2/2) + Log[a])^2)/b^2))/(a b) + (
       1.3734193849713405` E^(-((
         0.03333333333333333` (-4.605170185988092` + 
            15.` (0.05` - b^2/2) + Log[a])^2)/
         b^2)) (-4.605170185988092` + 15.` (0.05` - b^2/2) + 
          Log[a]))/(a^2 b^3) - (1/(a b^3))
       0.013734193849713406` E^(
        7.5` b^2 + 15.` (0.05` - b^2/2) - (
         0.03333333333333333` (-4.605170185988092` + 15.` b^2 + 
            15.` (0.05` - b^2/2) + Log[a])^2)/
         b^2) (-4.605170185988092` + 15.` b^2 + 15.` (0.05` - b^2/2) +
           Log[a])))