Use Implicit Differentiation to Find $\frac { \partial z } { \partial x }$

Use implicit differentiation to find $\frac { \partial z } { \partial x }$ $\ln \left( x ^ { 2 } + z ^ { 2 } \right) + y z ^ { 3 } + 4 x ^ { 2 } = 6$

A) $\frac { \partial z } { \partial x } = - \frac { 2 x \left( x ^ { 2 } + z ^ { 2 } + 6 \right) } { z \left( 3 y z ^ { 3 } + 3 x ^ { 2 } y z + 6 \right) }$
B) $\frac { \partial z } { \partial x } = - \frac { 2 x \left( 4 x ^ { 2 } + 4 z ^ { 2 } + 1 \right) } { z \left( 3 y z ^ { 3 } + 3 x ^ { 2 } y z + 2 \right) }$
C) $\frac { \partial z } { \partial x } = - \frac { 2 x \left( 4 x ^ { 2 } + 4 z ^ { 2 } + 6 \right) } { z \left( 3 y z ^ { 3 } + 3 x ^ { 2 } y z + 6 \right) }$
D) $\frac { \partial z } { \partial x } = - \frac { 2 x \left( 6 x ^ { 2 } + 6 z ^ { 2 } + 1 \right) } { z \left( 3 y z ^ { 3 } + 3 x ^ { 2 } y z + 2 \right) }$

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