3. Let $$ f(x, y)=\left\{\begin{array}{ll}\frac{x y\left(x^{2}-y^{2}\right)}{x^{2}+y^{2}}, & (x, y)=(0,0) \ 0, & (x, y) \neq(0,0)\end{array}\right. $$ Compute (a) $$ f_{x}(0,0) $$ (b) $$ f_{y}(0,0) $$ (c) $$ f_{x x}(0,0) $$ (d) $$ f_{y y}(0,0) $$ (e) $$ f_{x y}(0,0) $$ (f) $$ f_{y x}(0,0) $$

Question

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We are given
$$
f(x,y)=
\begin{cases}
\displaystyle \frac{xy\,(x^2-y^2)}{x^2+y^2}, & (x,y)\neq (0,0), \[1mm]
0, & (x,y)=(0,0).
\end{cases}
$$
We wish to compute the following partial derivatives at $$(0,0)$$:

(a) $$ f_{x}(0,0) $$  
(b) $$ f_{y}(0,0) $$  
(c) $$ f_{xx}(0,0) $$  
(d) $$ f_{yy}(0,0) $$  
(e) $$ f_{xy}(0,0) $$  
(f) $$ f_{yx}(0,0) $$

We work through each step-by-step.

---

**(a) Compute $$ f_{x}(0,0) $$:**

By definition,
$$
f_{x}(0,0)=\lim_{h\to 0}\frac{f(h,0)-f(0,0)}{h}.
$$
When $$y=0$$, note that
$$
f(h,0)=\frac{h\cdot 0\,(h^2-0^2)}{h^2+0^2}=0.
$$
Since $$f(0,0)=0$$, we have
$$
f_{x}(0,0)=\lim_{h\to 0}\frac{0-0}{h} = 0.
$$

---

**(b) Compute $$ f_{y}(0,0) $$:**

Similarly, by definition,
$$
f_{y}(0,0)=\lim_{k\to 0}\frac{f(0,k)-f(0,0)}{k}.
$$
When $$x=0$$,
$$
f(0,k)=\frac{0\cdot k\,(0^2-k^2)}{0+k^2}=0.
$$
Thus,
$$
f_{y}(0,0)=\lim_{k\to 0}\frac{0-0}{k} = 0.
$$

---

**(c) Compute $$ f_{xx}(0,0) $$:**

The second derivative with respect to $$x$$ is defined as
$$
f_{xx}(0,0)=\lim_{h\to 0}\frac{f_{x}(h,0)-f_{x}(0,0)}{h}.
$$
We already found $$f( x,0)=0$$ for all $$x$$, so
$$
f_{x}(x,0)=\lim_{s\to 0}\frac{f(x+s,0)-f(x,0)}{s}=\lim_{s\to 0}\frac{0-0}{s}=0.
$$
In particular, $$f_{x}(h,0)=0$$ and $$f_{x}(0,0)=0$$. Therefore,
$$
f_{xx}(0,0)=\lim_{h\to 0}\frac{0-0}{h} = 0.
$$

---

**(d) Compute $$ f_{yy}(0,0) $$:**

Similarly,
$$
f_{yy}(0,0)=\lim_{k\to 0}\frac{f_{y}(0,k)-f_{y}(0,0)}{k}.
$$
We argued that $$f(0,y)=0$$ for all $$y$$ so that $$f_{y}(0,k)=0$$ identically. Thus,
$$
f_{yy}(0,0)=\lim_{k\to 0}\frac{0-0}{k} = 0.
$$

---

**(e) Compute $$ f_{xy}(0,0) $$:**

This mixed derivative is given by
$$
f_{xy}(0,0)=\lim_{k\to 0}\frac{f_{x}(0,k)-f_{x}(0,0)}{k}.
$$
We already know $$f_{x}(0,0)=0$$. So we need to compute $$f_{x}(0,k)$$. By definition,
$$
f_{x}(0,k)=\lim_{h\to 0}\frac{f(h,k)-f(0,k)}{h}=\lim_{h\to 0}\frac{f(h,k)}{h},
$$
since $$f(0,k)=0$$ for any $$k$$.

For $$(h,k)\neq(0,0)$$,
$$
f(h,k)=\frac{hk\,(h^2-k^2)}{h^2+k^2}.
$$
Then,
$$
f_{x}(0,k)=\lim_{h\to 0}\frac{1}{h}\cdot \frac{hk\,(h^2-k^2)}{h^2+k^2}
=\lim_{h\to 0}\frac{k\,(h^2-k^2)}{h^2+k^2}.
$$
Now, taking the limit as $$h\to0$$:
- The numerator becomes $$k\,(0-k^2)=-k^3$$.
- The denominator becomes $$0+k^2=k^2$$.

Thus,
$$
f_{x}(0,k)=\frac{-k^3}{k^2}=-k.
$$
Finally,
$$
f_{xy}(0,0)=\lim_{k\to 0}\frac{-k-0}{k}=\lim_{k\to 0}(-1)=-1.
$$

---

**(f) Compute $$ f_{yx}(0,0) $$:**

We now compute
$$
f_{yx}(0,0)=\lim_{h\to 0}\frac{f_{y}(h,0)-f_{y}(0,0)}{h}.
$$
Since $$f_{y}(0,0)=0$$, we need $$f_{y}(h,0)$$. By definition,
$$
f_{y}(h,0)=\lim_{k\to 0}\frac{f(h,k)-f(h,0)}{k}=\lim_{k\to 0}\frac{f(h,k)}{k},
$$
because $$f(h,0)=0$$ for all $$h$$.

Using $$f(h,k)=\frac{hk\,(h^2-k^2)}{h^2+k^2}$$,
$$
f_{y}(h,0)=\lim_{k\to 0}\frac{1}{k}\cdot \frac{hk\,(h^2-k^2)}{h^2+k^2}
=\lim_{k\to 0}\frac{h\,(h^2-k^2)}{h^2+k^2}.
$$
Now, when $$k\to 0$$:
- The numerator becomes $$h\,(h^2-0)=h^3$$,
- The denominator becomes $$h^2+0 = h^2$$.

Thus,
$$
f_{y}(h,0)=\frac{h^3}{h^2}=h.
$$
Then,
$$
f_{yx}(0,0)=\lim_{h\to 0}\frac{h-0}{h}=\lim_{h\to 0}1=1.
$$

---

Summarizing, we have:
$$
\begin{array}{rl}
\text{(a)}\quad f_{x}(0,0) &= 0, \
\text{(b)}\quad f_{y}(0,0) &= 0, \
\text{(c)}\quad f_{xx}(0,0) &= 0, \
\text{(d)}\quad f_{yy}(0,0) &= 0, \
\text{(e)}\quad f_{xy}(0,0) &= -1, \
\text{(f)}\quad f_{yx}(0,0) &= 1.
\end{array}
$$
At (0,0): - $$ f_{x} = 0 $$ - $$ f_{y} = 0 $$ - $$ f_{xx} = 0 $$ - $$ f_{yy} = 0 $$ - $$ f_{xy} = -1 $$ - $$ f_{yx} = 1 $$

Let

Compute
(a)
(b)
©
(d)
(e)
(f)

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Step 2: Compute

Step 3: Compute

Step 4: Compute

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