<s>[INST] Prove that the function f(x,y) is not differentiable I have the following problem:
Prove that the function:
$f(x,y)=
\ \begin{cases} 
      \frac{x^3-x\cdot y^2}{x^2+y^2} & (x,y)\neq (0,0) \\
       \\0 & (x,y)=(0,0)
   \end{cases}
\\$
is continuous on $R^2$ and has its first order partial derivatives.
 everywhere on $R^2$, but $f$ is not differentiable at $(0,0)$
I know how to prove that it is continuous on $R^2$ and its partial derivatives exist at $(0,0)$ (I use limit definition of a derivative). But I do not know how to prove that this function is not differentiable.
 [/INST] We have $${ f }_{ x }^{ \prime  }\left( 0,0 \right) =\lim _{ x\rightarrow 0 }{ \frac { f\left( x,0 \right) -f\left( 0,0 \right)  }{ x }  } =\lim _{ x\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } }  }{ x }  } =1\\ \\ { f }_{ y }^{ \prime  }\left( 0,0 \right) =\lim _{ y\rightarrow 0 }{ \frac { f\left( 0,y \right) -f\left( 0,0 \right)  }{ y }  } =\lim _{ y\rightarrow 0 }{ \frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } }  }{ y }  } =0\\ f\left( x,y \right) -f\left( 0,0 \right) =\frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } =x+\left( \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x \right) ={ f }_{ x }^{ \prime  }\left( 0,0 \right) x+{ f }_{ y }^{ \prime  }\left( 0,0 \right) y+\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$
where $\alpha \left( x,y \right) =\frac { \frac { x^{ 3 }-x\cdot y^{ 2 } }{ x^{ 2 }+y^{ 2 } } -x }{ \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  } =\frac { -2x{ y }^{ 2 } }{ \left( x^{ 2 }+y^{ 2 } \right) \sqrt { x^{ 2 }+y^{ 2 } }  } $
however when $n\rightarrow \infty $ $$\alpha \left( \frac { 1 }{ n } ,\frac { 1 }{ n }  \right) =\frac { -\frac { 1 }{ { n }^{ 3 } }  }{ \frac { 1 }{ { n }^{ 3 } } \sqrt { 2 }  } =-\frac { 1 }{ \sqrt { 2 }  } $$
which shows $$\alpha \left( x,y \right) \sqrt { { x }^{ 2 }+{ y }^{ 2 } } \neq o\left( \sqrt { { x }^{ 2 }+{ y }^{ 2 } }  \right) $$
</s>