Correct Answer  Option 4 : f has no extreme value
Concept:
\(r = {\left( {\frac{{{\partial ^2}f}}{{\partial {x^2}}}} \right)_{\left( {a,b} \right)}},\;s = {\left( {\frac{{{\partial ^2}f}}{{\partial x\partial y}}} \right)_{\left( {a,b} \right)}},\;t = \;{\left( {\frac{{{\partial ^2}f}}{{\partial {y^2}}}} \right)_{\left( {a,b} \right)}}\)
 If f will be maximum or minimum, if rt – s^{2} is +ve → (> 0), s^{2} – rt is –ve → (<0)

Based on r → ve ⇒ f is maximum at (a, b)

Based on r → +ve ⇒ f is minimum at (a, b)
 If rt – s^{2} is –ve, (a, b) saddle point i.e. neither maximum nor minimum
Calculation:
Given:
\(f\left( {x,y} \right) = \;{y^2}  {x^2}\)
Differentiating with respect to f_{x},
2x = 0 ⇒ x = 0
Differentiating with respect to f_{y},
2y = 0 ⇒ y = 0
Again differentiating for second time, we get
r = f_{xx} = 2
s = f_{yy} = 2
Now, f_{xy} = 0 = s
\(rt  {s^2} = \;  4\;\left( {negative} \right)\)
∴ Neither maximum nor minimum