设\(z = z\left( {x,y} \right)\)是由方程\({z^3}{\rm{ + }}3xyz - 3\sin xy = 1\)确定的隐函数，则\( { { \partial z} \over {\partial y}}=\)( ) A: \( { { y\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) B: \( { { y\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\) C: \( { { x\left( {\cos xy - z} \right)} \over { { z^2} + xy}}\) D: \( { { x\left( {z - \cos xy} \right)} \over { { z^2} + xy}}\)

下列等式正确的有：{多选题} A: w'z' + w'xy + wx'z + wxyz = w'z' + xyz + wx'y'z + wyz B: z + y' + yz' = 1 C: xy'z' + x' + xyz' = x' + zy' D: xy + x'z + yz = xy + x'z

设方程\({e^z} - xyz = 0\)确定函数\(z=z(x,y)\)，则\( { { \partial z} \over {\partial x}}=\) A: \( { { yz} \over { { e^z} - xy}}\) B: \(- { { yz} \over { { e^z} - xy}}\) C: \( { { yz} \over { { e^z} +xy}}\) D: \(- { { yz} \over { { e^z}+xy}}\)

9. 已知函数$z=z(x,y)$由${{z}^{3}}-3xyz={{a}^{3}}$确定，则$\frac{{{\partial }^{2}}z}{\partial x\partial y}=$（ ） A: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ B: $\frac{z({{z}^{4}}-2xy{{z}^{2}}-xy)}{{{({{z}^{2}}-xy)}^{2}}}$ C: $\frac{z({{z}^{3}}-2xyz-{{x}^{2}}{{y}^{2}})}{{{({{z}^{2}}-xy)}^{3}}}$ D: $\frac{z({{z}^{3}}-2xy{{z}^{2}}-{{x}^{2}}y)}{{{({{z}^{2}}-xy)}^{3}}}$

由方程\({z^3} - 3xyz = {a^3}\)所确定的隐函数\(z= f(x,y)\)的偏导数\( { { \partial z} \over {\partial x}} = \) A: \( { { yz} \over { { z^2} - xy}}\) B: \(- { { yz} \over { { z^2} - xy}}\) C: \( { { yz} \over { { z^2} +xy}}\) D: \(- { { yz} \over { { z^2}+xy}}\)

函数z＝xy

设\(z = u{e^v}\)，\(u = x + y\)，\(v = xy\)，则\( { { \partial z} \over {\partial x}}=\) A: \({e^{xy}}(1 + xy + {y^2})\) B: \({e^{xy}}(1 + xy + {y^3})\) C: \({e^{xy}}(x+ xy + {y^2})\) D: \({e^{xy}}(y+ xy + {y^2})\)

设\(z = xy{e^{\sin xy}}\)，则\({z'_y} = \)（ ）。 A: \(x{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) B: \(y{e^{\sin xy}}\left( {1 + xy\cos xy} \right)\) C: \(x{e^{\sin xy}}\left( {1 + y\cos xy} \right)\) D: \(x{e^{\sin xy}}\left( {1 - xy\cos xy} \right)\)

设z=xy，则dz=______ A: yxy-1dx+xylnxdy B: xy-1dx+ydy C: xy(dx+dy) D: xy(xdx+ydy)

z=xy上原点为（）。