Use a Regra da Cadeia para determinar o que se pede:
R=\ln \left(u^{2}+v^{2}+w^{2}\right), u=x+y, v=2 x-y, w=2 x y ; \frac{\partial R}{\partial x}(1,1) e \,\frac{\partial R}{\partial y}(1,1).

Passo 1

O enunciado pede as derivadas parciais \frac{\partial R}{\partial x} e \frac{\partial R}{\partial y} no ponto (1,1) da função
R=\ln\bigl(u^{2}+v^{2}+w^{2}\bigr), \quad u=x+y,\;v=2x-y,\;w=2xy.
Vamos aplicar a Regra da Cadeia, pois R depende de u,v,w, que por sua vez dependem de x,y.

Passo 2

Para \frac{\partial R}{\partial x}, a Regra da Cadeia diz:
\frac{\partial R}{\partial x} = \frac{1}{u^{2}+v^{2}+w^{2}} \bigl(2u\,\frac{\partial u}{\partial x} + 2v\,\frac{\partial v}{\partial x} + 2w\,\frac{\partial w}{\partial x}\bigr).
Em (1,1) temos \bigl(u,v,w\bigr)=(2,1,2), \frac{\partial u}{\partial x}=1, \frac{\partial v}{\partial x}=2, \frac{\partial w}{\partial x}=2.
Logo,
\frac{\partial R}{\partial x}(1,1) = \frac{1}{4+1+4}\bigl(2\cdot2\cdot1 + 2\cdot1\cdot2 + 2\cdot2\cdot2\bigr) = \frac{16}{9}.

Passo 3

Para \frac{\partial R}{\partial y}, de modo análogo:
\frac{\partial R}{\partial y} = \frac{1}{u^{2}+v^{2}+w^{2}} \bigl(2u\,\frac{\partial u}{\partial y} + 2v\,\frac{\partial v}{\partial y} + 2w\,\frac{\partial w}{\partial y}\bigr).
Em (1,1) temos \frac{\partial u}{\partial y}=1, \frac{\partial v}{\partial y}=-1, \frac{\partial w}{\partial y}=2.
Logo,
\frac{\partial R}{\partial y}(1,1) = \frac{1}{9}\bigl(2\cdot2\cdot1 + 2\cdot1\cdot(-1) + 2\cdot2\cdot2\bigr) = \frac{10}{9}.

Resposta

\displaystyle \frac{\partial R}{\partial x}(1,1)=\frac{16}{9} \displaystyle \frac{\partial R}{\partial y}(1,1)=\frac{10}{9}

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