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?

What would be the best way(s) to tackle this in MMA?

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])))