Discuss continuity and differentiability of the function \(f : \mathbb { R } ^ { 2 } \to \mathbb { R }\) defined by
\[ f ( x , y ) = \left\{ \begin{array} { l l } { \frac { x y } { \sqrt { x ^ { 2 } + y ^ { 2 } } } , \qquad } & { ( x , y ) \neq ( 0 , 0 ) } \\ { 0 , } & { ( x , y ) = ( 0 , 0 ) } \end{array} . \right. \]
Solution
1. The function \(f\) is continuous at \((0,0)\).
For \((x,y)\neq(0,0)\), the inequality \[ 2\abs{xy}\leq x^2+y^2 \] gives \[ \abs{f(x,y)} = \frac{\abs{xy}}{\sqrt{x^2+y^2}} \leq \frac12\sqrt{x^2+y^2}. \] The right-hand side tends to \(0=f(0,0)\) as \((x,y)\to(0,0)\). Hence \(f\) is continuous at the origin.
2. The function \(f\) is differentiable at every\[ (x,y)\neq(0,0), \]and there\[ Df(x,y)(h,k) = \frac{y^3h+x^3k}{(x^2+y^2)^{3/2}}. \]
Away from the origin, the denominator \[ \sqrt{x^2+y^2} \] is nonzero, so \(f\) is a composition of smooth functions. Direct differentiation gives \[ \frac{\partial f}{\partial x} = \frac{y^3}{(x^2+y^2)^{3/2}}, \qquad \frac{\partial f}{\partial y} = \frac{x^3}{(x^2+y^2)^{3/2}}. \] Thus the total derivative has the displayed form.
3. Both first partial derivatives of \(f\) exist at \((0,0)\) and equal\(0\).
Along either coordinate axis, \(f\) is identically zero: \[ f(h,0)=0, \qquad f(0,k)=0. \] Therefore \[ \frac{\partial f}{\partial x}(0,0) = \lim_{h\to0}\frac{f(h,0)-f(0,0)}h = 0 \] and similarly \[ \frac{\partial f}{\partial y}(0,0)=0. \]
4. The function \(f\) is not differentiable at \((0,0)\).
If \(f\) were differentiable at the origin, step 3 would force its derivative there to be the zero linear map. Differentiability would then imply \[ \frac{\abs{f(h,k)}}{\sqrt{h^2+k^2}} \longrightarrow 0 \] as \((h,k)\to(0,0)\).
For \(t\neq0\), however, \[ f(t,t) = \frac{t^2}{\sqrt{2t^2}} = \frac{\abs t}{\sqrt2}, \] so \[ \frac{\abs{f(t,t)}}{\sqrt{t^2+t^2}} = \frac12. \] This does not tend to zero as \(t\to0\). Hence \(f\) is not differentiable at the origin.
5. Consequently,\[ \boxed{ f\text{ is continuous on }\RR^2 \text{ and differentiable exactly on }\RR^2\sm\{(0,0)\}. } \]
2. Step 2 proves differentiability off the origin, while step4 rules it out at the origin.
6. Q.E.D.
Step 5 gives the requested continuity and differentiability classification.